Mixing
$x^y = y^x$ for integers $x$ and $y$
$begingroup$
We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($xneq y$) which satisfies the equality $x^y = y^x$?
number-theory diophantine-equations exponentiation
edited Dec 4 '11 at 13:25
J. M. is not a mathematician
61.2k5152290
asked Nov 9 '10 at 0:23
Paulo ArgoloPaulo Argolo
1,45332336
$endgroup$
add a comment |
$begingroup$
We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($xneq y$) which satisfies the equality $x^y = y^x$?
number-theory diophantine-equations exponentiation
edited Dec 4 '11 at 13:25
J. M. is not a mathematician
61.2k5152290
asked Nov 9 '10 at 0:23
Paulo ArgoloPaulo Argolo
1,45332336
$endgroup$
add a comment |
$begingroup$
We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($xneq y$) which satisfies the equality $x^y = y^x$?
number-theory diophantine-equations exponentiation
edited Dec 4 '11 at 13:25
J. M. is not a mathematician
61.2k5152290
asked Nov 9 '10 at 0:23
Paulo ArgoloPaulo Argolo
1,45332336
$endgroup$
We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($xneq y$) which satisfies the equality $x^y = y^x$?
number-theory diophantine-equations exponentiation
number-theory diophantine-equations exponentiation
edited Dec 4 '11 at 13:25
J. M. is not a mathematician
61.2k5152290
asked Nov 9 '10 at 0:23
Paulo ArgoloPaulo Argolo
1,45332336
edited Dec 4 '11 at 13:25
J. M. is not a mathematician
61.2k5152290
asked Nov 9 '10 at 0:23
Paulo ArgoloPaulo Argolo
1,45332336
edited Dec 4 '11 at 13:25
J. M. is not a mathematician
61.2k5152290
edited Dec 4 '11 at 13:25
J. M. is not a mathematician
61.2k5152290
edited Dec 4 '11 at 13:25
J. M. is not a mathematician
61.2k5152290
61.2k5152290
asked Nov 9 '10 at 0:23
Paulo ArgoloPaulo Argolo
1,45332336
asked Nov 9 '10 at 0:23
Paulo ArgoloPaulo Argolo
1,45332336
asked Nov 9 '10 at 0:23
Paulo ArgoloPaulo Argolo
1,45332336
1,45332336
add a comment |
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
This is a classic (and well known problem).
The general solution of $x^y = y^x$ is given by
$$begin{align*}x &= (1+1/u)^u \ y &= (1+1/u)^{u+1}end{align*}$$
It can be shown that if $x$ and $y$ are rational, then $u$ must be an integer.
For more details, see this and this.
edited Oct 31 '11 at 9:12
J. M. is not a mathematician
61.2k5152290
answered Nov 9 '10 at 1:40
AryabhataAryabhata
70.2k6157247
$endgroup$
9
$begingroup$
I know about this curve, but I haven't seen that parametrization before. +1 for that!
$endgroup$
– J. M. is not a mathematician
Nov 9 '10 at 2:19
4
$begingroup$
The parametrization, it turns out, is due to Christian Goldbach, which he did in 1843. (See this for instance.)
$endgroup$
– J. M. is not a mathematician
Dec 4 '11 at 13:20
$begingroup$
@J.M.: Thanks..!
$endgroup$
– Aryabhata
Dec 20 '11 at 20:58
$begingroup$
Dude, long time no see! Sure hope you're fine.
$endgroup$
– J. M. is not a mathematician
Dec 20 '11 at 23:52
6
$begingroup$
interesting... as u goes to infinity, x and y go to e
$endgroup$
– Mehrdad
Apr 8 '13 at 19:11
|
show 5 more comments
$begingroup$
For every integer $n$, $x = y = n$ is a solution. So assume $x neq y$.
Suppose $n^m = m^n$. Then $n^{1/n} = m^{1/m}$. Now the function $x mapsto x^{1/x}$ reaches its maximum at $e$, and is otherwise monotone. Thus (assuming $n < m$) we must have $n < e$, i.e. $n = 1$ or $n = 2$.
If $n = 1$ then $n^m = 1$ and so $m = 1$, so it's a trivial solution.
If $n = 2$ then $n^m$ is a power of $2$, and so (since $m > 0$) $m$ must also be a power of $2$, say $m = 2^k$. Then $n^m = 2^{2^k}$ and $m^n = 2^{2k}$, so that $2^k = 2k$ or $2^{k-1} = k$. Now $2^{3-1} > 3$, and so an easy induction shows that $k leq 2$. If $k = 1$ then $n = m$, and $k = 2$ corresponds to $2^4 = 4^2$.
EDIT: Up till now we considered $n,m>0$. We now go over all other cases. The solution $n = m = 0$ is trivial (whatever value we give to $0^0$).
If $n=0$ and $m neq 0$ then $n^m = 0$ whereas $m^n = 1$, so this is not a solution.
If $n > 0$ and $m < 0$ then $0 < n^m leq 1$ whereas $|m^n| geq 1$. Hence necessarily $n^m = 1$ so that $n = 1$. It follows that $m^1 = 1^m = 1$. In particular, there's no solution with opposite signs.
If $n,m < 0$ then $(-1)^m (-n)^m = n^m = m^n = (-1)^n (-m)^n$, so that $n,m$ must have the same parity. Taking inverses, we get $(-n)^{-m} = (-m)^{-n}$, so that $-n,-m$ is a solution for positive integers. The only non-trivial positive solution $2,4$ yields the only non-trivial negative solution $-2,-4$.
answered Nov 9 '10 at 0:34
Yuval FilmusYuval Filmus
49k472148
$endgroup$
$begingroup$
I edited the question.
$endgroup$
– Paulo Argolo
Nov 9 '10 at 1:12
2
$begingroup$
@Paulo: Yuval answered your question adequately. He showed that the pair $(2,4)$ are the only distinct integers which satisfy, and that any pair $(x,x), x in mathbb{N}$ is a solution.
$endgroup$
– Brandon Carter
Nov 9 '10 at 1:17
$begingroup$
@Brandon Carter: (-2)^(-4)= (-4)(-2). I just edit the question. Thank you for the comment.
$endgroup$
– Paulo Argolo
Nov 9 '10 at 1:22
add a comment |
$begingroup$
Say $x^y = y^x$, and $x > y > 0$. Taking logs, $y log x = x log y$; rearranging, $(log x)/x = (log y)/y$. Let $f(x) = (log x)/x$; then this is $f(x) = f(y)$.
Now, $f^prime(x) = (1-log x)/x^2$, so $f$ is increasing for $x<e$ and decreasing for $x>e$. So if $x^y = y^x$ has a solution, then $x > e > y$. So $y$ must be $1$ or $2$. But $y = 1$ doesn't work. $y=2$ gives $x=4$.
(I've always thought of this as the ``standard'' solution to this problem and I'm a bit surprised nobody has posted it yet.)
If $x > 0 > y$, then $0 < x^y < 1$ and $y^x$ is an integer, so there are no such solutions.
If $0 > x > y$, then $x^y = y^x$ implies $x$ and $y$ must have the same parity. Also, taking reciprocals, $x^{-y} = y^{-x}$. Then $(-x)^{-y} = (-y)^{-x}$ since $x$ and $y$ have the same parity. (The number of factors of $-1$ we introduced to each side differs by $x-y$, which is even.) So solving the problem where $x$ and $y$ are negative reduces to solving it when $x$ and $y$ are positive.
answered Nov 19 '10 at 19:49
Michael LugoMichael Lugo
18.4k33576
$endgroup$
add a comment |
$begingroup$
Although this thing has already been answered, here a shorter proof
Because $x^y = y^x $ is symmetric we first demand that $x>y$
Then we proceed simply this way:
$ x^y = y^x $
$ x = y^{frac x y } $
$ frac x y = y^{frac x y -1} $
$ frac x y -1 = y^{frac x y -1} - 1 $
Now we expand the rhs into its well-known exponential-series
$ frac x y -1 = ln(y)*(frac x y -1) + frac {((ln(y)*(frac x y -1))^2}{2!} + ... $
Here by the definition x>y the lhs is positive, so if $ ln(y) $ >=1 we had lhs $lt$ rhs
Thus $ ln(y) $ must be smaller than 1, and the only integer y>1 whose log is smaller than 1 is y=2, so there is the only possibility $y = 2$ and we are done.
[update] Well, after having determined $y=2$ the same procedure can be used to show, that after manyally checking $x=3$ (impossible) $x=4$ (possible) no $x>4$ can be chosen.
We ask for $x=4^{1+delta} ,delta > 0 $ inserting the value 2 for y:
$ 4^{(1+delta)*2}=2^{4^{(1+delta)}} $
Take log to base 2:
$ (1+delta)*4=4^{(1+delta)} $
$ delta =4^{delta} - 1 $
$ delta = ln(4)*delta + frac { (ln(4)*delta)^2 }{2!} + ldots $
$ 0 = (ln(4)-1)*delta + frac { (ln(4)*delta)^2 }{2!} + ldots $
Because $ ln(4)-1 >0 $ this can only be satisfied if $ delta =0 $
So indeed the only solutions, assuming x>y, is $ (x,y) = (4,2)$ .
[end update]
answered Nov 19 '10 at 17:02
Gottfried HelmsGottfried Helms
23.7k245101
$endgroup$
add a comment |
$begingroup$
Well I finally found an answer relating to some number theory I suppose !
Assume that : $x={p_1}^{alpha _ 1}.{p_2}^{alpha _ 2}...{p_k}^{alpha _ k}$ it is clear that number y prime factors are the same as number x but with different powers i.e: $y={p_1}^{beta _ 1}.{p_2}^{beta _ 2}...{p_k}^{beta _ k}$ replacing the first equation we get:
${({p_1}^{alpha _ 1}.{p_2}^{alpha _ 2}...{p_k}^{alpha _ k})}^y={({p_1}^{beta _ 1}.{p_2}^{beta _ 2}...{p_k}^{beta _ k})}^x$ i.e: ${p_1}^{{alpha _ 1}y}.{p_2}^{{alpha _ 2}y}...{p_k}^{{alpha _ k}y}={p_1}^{{beta _ 1}x}.{p_2}^{{beta _ 2}x}...{p_k}^{{beta _ k}x}$
Since the the powers ought to be equal we know for each $1le i le k$ we have:${alpha_i}y={beta_i}x$ i.e: ${alpha_i}/{beta_i}=x/y$
Considering that the equation is symmetric we can assume that $x le y$ but we have ${alpha_i}/{beta_i} = x/y ge 1$ hence ${alpha_i} ge {beta_i}$
Assume this obvious,easy-to-prove theorem:
Theorem #1
Consider $x,y in mathbb{N}$ such that $x={p_1}^{alpha _ 1}.{p_2}^{alpha _ 2}...{p_k}^{alpha _ k}$ $y={p_1}^{beta _ 1}.{p_2}^{beta _ 2}...{p_k}^{beta _ k}$ for each $1le i le k$ we have:
$y|x to {alpha_i}ge{beta_i}$ or vice versa
Using the Theorem #1 we can get that $y|x$ i.e $x=yt$ replacing in the main equation we get:
$x^y=y^x to ({yt})^y=y^{({yt})} to yt=y^t$
Now we must find the answers to the equation $yt=y^t$ for $t=1$ it is obvious that for every $y in mathbb{N}$ the equation is valid.so one answer is $x=y$
Yet again for $t=2$ we must have $2y=y^2$ i.e $y=2$ and we can conclude that $x=4$ (using the equation $x=yt$)so another answer is $x=4$ $land$ y=2$ (or vice versa)
We show that for $tge3$ the equation is not valid anymore.
If $tge3$ then $ygt2$ we prove that with these terms the inequality $y^t gt yt$ stands.
$y^t={(y-1+1)}^t={(y-1)}^t+...+binom{t}{2} {(y-1)}^2 + binom{t}{1}(y-1) +1 gt binom{t}{2} {(y-1)}^2 + t(y-1) +1$
But we have $y-1gt1$ so:
$y^t gt binom{t}{2} {(y-1)}^2 + t(y-1) +1= frac {t(t-1)}{2} -t +1 +yt= frac {(t-2)(t-1)}{2} + yt gt yt$
So it is proved that for $tge3$ is not valid anymore.$bullet$
P.S: The equation is solved for positive integers yet the solution for all the integers is quite the same!(took me an hour to write this all,hope you like my solution)
answered Jul 27 '14 at 6:33
FuriousMathematicianFuriousMathematician
320214
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add a comment |
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I've collected some references, feel free to add more of them. (Some of them are taken from other answers. And, of course, some of them can contain further interesting references.)
Online:
Nick's Mathematical Puzzles - Solution to puzzle 48: Exponential equation (Wayback Machine)
On Torsten Sillke's page: http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/x%5Ey-x%5Ey (Wayback Machine)
Wikipedia: Equation $x^y=y^x$
Papers:
Michael A. Bennett and Bruce Reznick: Positive Rational Solutions to $x^y = y^{mx}$ : A Number-Theoretic Excursion, The American Mathematical Monthly , Vol. 111, No. 1 (Jan., 2004), pp. 13-21; available at jstor, arxiv or at author's homepage.
Marta Sved: On the Rational Solutions of $x^y = y^x$, Mathematics Magazine, Vol. 63, No. 1 (Feb., 1990), pp. 30-33, available at jstor. It is mentioned here, that this problem appeared in 1960 Putnam Competition (for integers)
F. Gerrish: 76.25 $a^{b}=b^{a}$: The Positive Integer Solution, The Mathematical Gazette, Vol. 76, No. 477 (Nov., 1992), p. 403. Jstor link
Solomon Hurwitz: On the Rational Solutions of $m^n=n^m$ with $mne n$, The American Mathematical Monthly, Vol. 74, No. 3 (Mar., 1967), pp. 298-300. jstor
Joel Anderson: Iterated Exponentials, The American Mathematical Monthly , Vol. 111, No. 8 (Oct., 2004), pp. 668-679. jstor
R. Arthur Knoebel, Exponentials Reiterated. The American Mathematical Monthly , Vol. 88, No. 4 (Apr., 1981), pp. 235-252 jstor, link
Books:
Andrew M. Gleason, R. E. Greenwood, Leroy Milton Kelly: William Lowell Putnam mathematical competition problems and solutions 1938-1964, ISBN 0883854287, p.59 and p.538
Titu Andreescu, Dorin Andrica, Ion Cucurezeanu: An Introduction to Diophantine Equations: A Problem-Based Approach, Springer, New York, 2010. Page 209
Searches:
The reason I've added this is that it can be somewhat tricky to search for a formula or an equation. So any interesting idea which could help finding interesting references may be of interest.
- diophantine "x^y=y^x"
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add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a classic (and well known problem).
The general solution of $x^y = y^x$ is given by
$$begin{align*}x &= (1+1/u)^u \ y &= (1+1/u)^{u+1}end{align*}$$
It can be shown that if $x$ and $y$ are rational, then $u$ must be an integer.
For more details, see this and this.
edited Oct 31 '11 at 9:12
J. M. is not a mathematician
61.2k5152290
answered Nov 9 '10 at 1:40
AryabhataAryabhata
70.2k6157247
$endgroup$
9
$begingroup$
I know about this curve, but I haven't seen that parametrization before. +1 for that!
$endgroup$
– J. M. is not a mathematician
Nov 9 '10 at 2:19
4
$begingroup$
The parametrization, it turns out, is due to Christian Goldbach, which he did in 1843. (See this for instance.)
$endgroup$
– J. M. is not a mathematician
Dec 4 '11 at 13:20
$begingroup$
@J.M.: Thanks..!
$endgroup$
– Aryabhata
Dec 20 '11 at 20:58
$begingroup$
Dude, long time no see! Sure hope you're fine.
$endgroup$
– J. M. is not a mathematician
Dec 20 '11 at 23:52
6
$begingroup$
interesting... as u goes to infinity, x and y go to e
$endgroup$
– Mehrdad
Apr 8 '13 at 19:11
|
show 5 more comments
$begingroup$
This is a classic (and well known problem).
The general solution of $x^y = y^x$ is given by
$$begin{align*}x &= (1+1/u)^u \ y &= (1+1/u)^{u+1}end{align*}$$
It can be shown that if $x$ and $y$ are rational, then $u$ must be an integer.
For more details, see this and this.
edited Oct 31 '11 at 9:12
J. M. is not a mathematician
61.2k5152290
answered Nov 9 '10 at 1:40
AryabhataAryabhata
70.2k6157247
$endgroup$
9
$begingroup$
I know about this curve, but I haven't seen that parametrization before. +1 for that!
$endgroup$
– J. M. is not a mathematician
Nov 9 '10 at 2:19
4
$begingroup$
The parametrization, it turns out, is due to Christian Goldbach, which he did in 1843. (See this for instance.)
$endgroup$
– J. M. is not a mathematician
Dec 4 '11 at 13:20
$begingroup$
@J.M.: Thanks..!
$endgroup$
– Aryabhata
Dec 20 '11 at 20:58
$begingroup$
Dude, long time no see! Sure hope you're fine.
$endgroup$
– J. M. is not a mathematician
Dec 20 '11 at 23:52
6
$begingroup$
interesting... as u goes to infinity, x and y go to e
$endgroup$
– Mehrdad
Apr 8 '13 at 19:11
|
show 5 more comments
$begingroup$
This is a classic (and well known problem).
The general solution of $x^y = y^x$ is given by
$$begin{align*}x &= (1+1/u)^u \ y &= (1+1/u)^{u+1}end{align*}$$
It can be shown that if $x$ and $y$ are rational, then $u$ must be an integer.
For more details, see this and this.
edited Oct 31 '11 at 9:12
J. M. is not a mathematician
61.2k5152290
answered Nov 9 '10 at 1:40
AryabhataAryabhata
70.2k6157247
$endgroup$
This is a classic (and well known problem).
The general solution of $x^y = y^x$ is given by
$$begin{align*}x &= (1+1/u)^u \ y &= (1+1/u)^{u+1}end{align*}$$
It can be shown that if $x$ and $y$ are rational, then $u$ must be an integer.
For more details, see this and this.
edited Oct 31 '11 at 9:12
J. M. is not a mathematician
61.2k5152290
answered Nov 9 '10 at 1:40
AryabhataAryabhata
70.2k6157247
edited Oct 31 '11 at 9:12
J. M. is not a mathematician
61.2k5152290
edited Oct 31 '11 at 9:12
J. M. is not a mathematician
61.2k5152290
edited Oct 31 '11 at 9:12
J. M. is not a mathematician
61.2k5152290
61.2k5152290
answered Nov 9 '10 at 1:40
AryabhataAryabhata
70.2k6157247
answered Nov 9 '10 at 1:40
AryabhataAryabhata
70.2k6157247
answered Nov 9 '10 at 1:40
AryabhataAryabhata
70.2k6157247
70.2k6157247
9
$begingroup$
I know about this curve, but I haven't seen that parametrization before. +1 for that!
$endgroup$
– J. M. is not a mathematician
Nov 9 '10 at 2:19
4
$begingroup$
The parametrization, it turns out, is due to Christian Goldbach, which he did in 1843. (See this for instance.)
$endgroup$
– J. M. is not a mathematician
Dec 4 '11 at 13:20
$begingroup$
@J.M.: Thanks..!
$endgroup$
– Aryabhata
Dec 20 '11 at 20:58
$begingroup$
Dude, long time no see! Sure hope you're fine.
$endgroup$
– J. M. is not a mathematician
Dec 20 '11 at 23:52
6
$begingroup$
interesting... as u goes to infinity, x and y go to e
$endgroup$
– Mehrdad
Apr 8 '13 at 19:11
|
show 5 more comments
9
$begingroup$
I know about this curve, but I haven't seen that parametrization before. +1 for that!
$endgroup$
– J. M. is not a mathematician
Nov 9 '10 at 2:19
4
$begingroup$
The parametrization, it turns out, is due to Christian Goldbach, which he did in 1843. (See this for instance.)
$endgroup$
– J. M. is not a mathematician
Dec 4 '11 at 13:20
$begingroup$
@J.M.: Thanks..!
$endgroup$
– Aryabhata
Dec 20 '11 at 20:58
$begingroup$
Dude, long time no see! Sure hope you're fine.
$endgroup$
– J. M. is not a mathematician
Dec 20 '11 at 23:52
6
$begingroup$
interesting... as u goes to infinity, x and y go to e
$endgroup$
– Mehrdad
Apr 8 '13 at 19:11
9
9
$begingroup$
I know about this curve, but I haven't seen that parametrization before. +1 for that!
$endgroup$
– J. M. is not a mathematician
Nov 9 '10 at 2:19
$begingroup$
I know about this curve, but I haven't seen that parametrization before. +1 for that!
$endgroup$
– J. M. is not a mathematician
Nov 9 '10 at 2:19
4
4
$begingroup$
The parametrization, it turns out, is due to Christian Goldbach, which he did in 1843. (See this for instance.)
$endgroup$
– J. M. is not a mathematician
Dec 4 '11 at 13:20
$begingroup$
The parametrization, it turns out, is due to Christian Goldbach, which he did in 1843. (See this for instance.)
$endgroup$
– J. M. is not a mathematician
Dec 4 '11 at 13:20
$begingroup$
@J.M.: Thanks..!
$endgroup$
– Aryabhata
Dec 20 '11 at 20:58
$begingroup$
@J.M.: Thanks..!
$endgroup$
– Aryabhata
Dec 20 '11 at 20:58
$begingroup$
Dude, long time no see! Sure hope you're fine.
$endgroup$
– J. M. is not a mathematician
Dec 20 '11 at 23:52
$begingroup$
Dude, long time no see! Sure hope you're fine.
$endgroup$
– J. M. is not a mathematician
Dec 20 '11 at 23:52
6
6
$begingroup$
interesting... as u goes to infinity, x and y go to e
$endgroup$
– Mehrdad
Apr 8 '13 at 19:11
$begingroup$
interesting... as u goes to infinity, x and y go to e
$endgroup$
– Mehrdad
Apr 8 '13 at 19:11
|
show 5 more comments
$begingroup$
For every integer $n$, $x = y = n$ is a solution. So assume $x neq y$.
Suppose $n^m = m^n$. Then $n^{1/n} = m^{1/m}$. Now the function $x mapsto x^{1/x}$ reaches its maximum at $e$, and is otherwise monotone. Thus (assuming $n < m$) we must have $n < e$, i.e. $n = 1$ or $n = 2$.
If $n = 1$ then $n^m = 1$ and so $m = 1$, so it's a trivial solution.
If $n = 2$ then $n^m$ is a power of $2$, and so (since $m > 0$) $m$ must also be a power of $2$, say $m = 2^k$. Then $n^m = 2^{2^k}$ and $m^n = 2^{2k}$, so that $2^k = 2k$ or $2^{k-1} = k$. Now $2^{3-1} > 3$, and so an easy induction shows that $k leq 2$. If $k = 1$ then $n = m$, and $k = 2$ corresponds to $2^4 = 4^2$.
EDIT: Up till now we considered $n,m>0$. We now go over all other cases. The solution $n = m = 0$ is trivial (whatever value we give to $0^0$).
If $n=0$ and $m neq 0$ then $n^m = 0$ whereas $m^n = 1$, so this is not a solution.
If $n > 0$ and $m < 0$ then $0 < n^m leq 1$ whereas $|m^n| geq 1$. Hence necessarily $n^m = 1$ so that $n = 1$. It follows that $m^1 = 1^m = 1$. In particular, there's no solution with opposite signs.
If $n,m < 0$ then $(-1)^m (-n)^m = n^m = m^n = (-1)^n (-m)^n$, so that $n,m$ must have the same parity. Taking inverses, we get $(-n)^{-m} = (-m)^{-n}$, so that $-n,-m$ is a solution for positive integers. The only non-trivial positive solution $2,4$ yields the only non-trivial negative solution $-2,-4$.
answered Nov 9 '10 at 0:34
Yuval FilmusYuval Filmus
49k472148
$endgroup$
$begingroup$
I edited the question.
$endgroup$
– Paulo Argolo
Nov 9 '10 at 1:12
2
$begingroup$
@Paulo: Yuval answered your question adequately. He showed that the pair $(2,4)$ are the only distinct integers which satisfy, and that any pair $(x,x), x in mathbb{N}$ is a solution.
$endgroup$
– Brandon Carter
Nov 9 '10 at 1:17
$begingroup$
@Brandon Carter: (-2)^(-4)= (-4)(-2). I just edit the question. Thank you for the comment.
$endgroup$
– Paulo Argolo
Nov 9 '10 at 1:22
add a comment |
$begingroup$
For every integer $n$, $x = y = n$ is a solution. So assume $x neq y$.
Suppose $n^m = m^n$. Then $n^{1/n} = m^{1/m}$. Now the function $x mapsto x^{1/x}$ reaches its maximum at $e$, and is otherwise monotone. Thus (assuming $n < m$) we must have $n < e$, i.e. $n = 1$ or $n = 2$.
If $n = 1$ then $n^m = 1$ and so $m = 1$, so it's a trivial solution.
If $n = 2$ then $n^m$ is a power of $2$, and so (since $m > 0$) $m$ must also be a power of $2$, say $m = 2^k$. Then $n^m = 2^{2^k}$ and $m^n = 2^{2k}$, so that $2^k = 2k$ or $2^{k-1} = k$. Now $2^{3-1} > 3$, and so an easy induction shows that $k leq 2$. If $k = 1$ then $n = m$, and $k = 2$ corresponds to $2^4 = 4^2$.
EDIT: Up till now we considered $n,m>0$. We now go over all other cases. The solution $n = m = 0$ is trivial (whatever value we give to $0^0$).
If $n=0$ and $m neq 0$ then $n^m = 0$ whereas $m^n = 1$, so this is not a solution.
If $n > 0$ and $m < 0$ then $0 < n^m leq 1$ whereas $|m^n| geq 1$. Hence necessarily $n^m = 1$ so that $n = 1$. It follows that $m^1 = 1^m = 1$. In particular, there's no solution with opposite signs.
If $n,m < 0$ then $(-1)^m (-n)^m = n^m = m^n = (-1)^n (-m)^n$, so that $n,m$ must have the same parity. Taking inverses, we get $(-n)^{-m} = (-m)^{-n}$, so that $-n,-m$ is a solution for positive integers. The only non-trivial positive solution $2,4$ yields the only non-trivial negative solution $-2,-4$.
answered Nov 9 '10 at 0:34
Yuval FilmusYuval Filmus
49k472148
$endgroup$
$begingroup$
I edited the question.
$endgroup$
– Paulo Argolo
Nov 9 '10 at 1:12
2
$begingroup$
@Paulo: Yuval answered your question adequately. He showed that the pair $(2,4)$ are the only distinct integers which satisfy, and that any pair $(x,x), x in mathbb{N}$ is a solution.
$endgroup$
– Brandon Carter
Nov 9 '10 at 1:17
$begingroup$
@Brandon Carter: (-2)^(-4)= (-4)(-2). I just edit the question. Thank you for the comment.
$endgroup$
– Paulo Argolo
Nov 9 '10 at 1:22
add a comment |
$begingroup$
For every integer $n$, $x = y = n$ is a solution. So assume $x neq y$.
Suppose $n^m = m^n$. Then $n^{1/n} = m^{1/m}$. Now the function $x mapsto x^{1/x}$ reaches its maximum at $e$, and is otherwise monotone. Thus (assuming $n < m$) we must have $n < e$, i.e. $n = 1$ or $n = 2$.
If $n = 1$ then $n^m = 1$ and so $m = 1$, so it's a trivial solution.
If $n = 2$ then $n^m$ is a power of $2$, and so (since $m > 0$) $m$ must also be a power of $2$, say $m = 2^k$. Then $n^m = 2^{2^k}$ and $m^n = 2^{2k}$, so that $2^k = 2k$ or $2^{k-1} = k$. Now $2^{3-1} > 3$, and so an easy induction shows that $k leq 2$. If $k = 1$ then $n = m$, and $k = 2$ corresponds to $2^4 = 4^2$.
EDIT: Up till now we considered $n,m>0$. We now go over all other cases. The solution $n = m = 0$ is trivial (whatever value we give to $0^0$).
If $n=0$ and $m neq 0$ then $n^m = 0$ whereas $m^n = 1$, so this is not a solution.
If $n > 0$ and $m < 0$ then $0 < n^m leq 1$ whereas $|m^n| geq 1$. Hence necessarily $n^m = 1$ so that $n = 1$. It follows that $m^1 = 1^m = 1$. In particular, there's no solution with opposite signs.
If $n,m < 0$ then $(-1)^m (-n)^m = n^m = m^n = (-1)^n (-m)^n$, so that $n,m$ must have the same parity. Taking inverses, we get $(-n)^{-m} = (-m)^{-n}$, so that $-n,-m$ is a solution for positive integers. The only non-trivial positive solution $2,4$ yields the only non-trivial negative solution $-2,-4$.
answered Nov 9 '10 at 0:34
Yuval FilmusYuval Filmus
49k472148
$endgroup$
For every integer $n$, $x = y = n$ is a solution. So assume $x neq y$.
Suppose $n^m = m^n$. Then $n^{1/n} = m^{1/m}$. Now the function $x mapsto x^{1/x}$ reaches its maximum at $e$, and is otherwise monotone. Thus (assuming $n < m$) we must have $n < e$, i.e. $n = 1$ or $n = 2$.
If $n = 1$ then $n^m = 1$ and so $m = 1$, so it's a trivial solution.
If $n = 2$ then $n^m$ is a power of $2$, and so (since $m > 0$) $m$ must also be a power of $2$, say $m = 2^k$. Then $n^m = 2^{2^k}$ and $m^n = 2^{2k}$, so that $2^k = 2k$ or $2^{k-1} = k$. Now $2^{3-1} > 3$, and so an easy induction shows that $k leq 2$. If $k = 1$ then $n = m$, and $k = 2$ corresponds to $2^4 = 4^2$.
EDIT: Up till now we considered $n,m>0$. We now go over all other cases. The solution $n = m = 0$ is trivial (whatever value we give to $0^0$).
If $n=0$ and $m neq 0$ then $n^m = 0$ whereas $m^n = 1$, so this is not a solution.
If $n > 0$ and $m < 0$ then $0 < n^m leq 1$ whereas $|m^n| geq 1$. Hence necessarily $n^m = 1$ so that $n = 1$. It follows that $m^1 = 1^m = 1$. In particular, there's no solution with opposite signs.
If $n,m < 0$ then $(-1)^m (-n)^m = n^m = m^n = (-1)^n (-m)^n$, so that $n,m$ must have the same parity. Taking inverses, we get $(-n)^{-m} = (-m)^{-n}$, so that $-n,-m$ is a solution for positive integers. The only non-trivial positive solution $2,4$ yields the only non-trivial negative solution $-2,-4$.
answered Nov 9 '10 at 0:34
Yuval FilmusYuval Filmus
49k472148
edited Nov 16 '10 at 18:28
answered Nov 9 '10 at 0:34
Yuval FilmusYuval Filmus
49k472148
answered Nov 9 '10 at 0:34
Yuval FilmusYuval Filmus
49k472148
answered Nov 9 '10 at 0:34
Yuval FilmusYuval Filmus
49k472148
49k472148
$begingroup$
I edited the question.
$endgroup$
– Paulo Argolo
Nov 9 '10 at 1:12
2
$begingroup$
@Paulo: Yuval answered your question adequately. He showed that the pair $(2,4)$ are the only distinct integers which satisfy, and that any pair $(x,x), x in mathbb{N}$ is a solution.
$endgroup$
– Brandon Carter
Nov 9 '10 at 1:17
$begingroup$
@Brandon Carter: (-2)^(-4)= (-4)(-2). I just edit the question. Thank you for the comment.
$endgroup$
– Paulo Argolo
Nov 9 '10 at 1:22
add a comment |
$begingroup$
I edited the question.
$endgroup$
– Paulo Argolo
Nov 9 '10 at 1:12
2
$begingroup$
@Paulo: Yuval answered your question adequately. He showed that the pair $(2,4)$ are the only distinct integers which satisfy, and that any pair $(x,x), x in mathbb{N}$ is a solution.
$endgroup$
– Brandon Carter
Nov 9 '10 at 1:17
$begingroup$
@Brandon Carter: (-2)^(-4)= (-4)(-2). I just edit the question. Thank you for the comment.
$endgroup$
– Paulo Argolo
Nov 9 '10 at 1:22
$begingroup$
I edited the question.
$endgroup$
– Paulo Argolo
Nov 9 '10 at 1:12
$begingroup$
I edited the question.
$endgroup$
– Paulo Argolo
Nov 9 '10 at 1:12
2
2
$begingroup$
@Paulo: Yuval answered your question adequately. He showed that the pair $(2,4)$ are the only distinct integers which satisfy, and that any pair $(x,x), x in mathbb{N}$ is a solution.
$endgroup$
– Brandon Carter
Nov 9 '10 at 1:17
$begingroup$
@Paulo: Yuval answered your question adequately. He showed that the pair $(2,4)$ are the only distinct integers which satisfy, and that any pair $(x,x), x in mathbb{N}$ is a solution.
$endgroup$
– Brandon Carter
Nov 9 '10 at 1:17
$begingroup$
@Brandon Carter: (-2)^(-4)= (-4)(-2). I just edit the question. Thank you for the comment.
$endgroup$
– Paulo Argolo
Nov 9 '10 at 1:22
$begingroup$
@Brandon Carter: (-2)^(-4)= (-4)(-2). I just edit the question. Thank you for the comment.
$endgroup$
– Paulo Argolo
Nov 9 '10 at 1:22
add a comment |
$begingroup$
Say $x^y = y^x$, and $x > y > 0$. Taking logs, $y log x = x log y$; rearranging, $(log x)/x = (log y)/y$. Let $f(x) = (log x)/x$; then this is $f(x) = f(y)$.
Now, $f^prime(x) = (1-log x)/x^2$, so $f$ is increasing for $x<e$ and decreasing for $x>e$. So if $x^y = y^x$ has a solution, then $x > e > y$. So $y$ must be $1$ or $2$. But $y = 1$ doesn't work. $y=2$ gives $x=4$.
(I've always thought of this as the ``standard'' solution to this problem and I'm a bit surprised nobody has posted it yet.)
If $x > 0 > y$, then $0 < x^y < 1$ and $y^x$ is an integer, so there are no such solutions.
If $0 > x > y$, then $x^y = y^x$ implies $x$ and $y$ must have the same parity. Also, taking reciprocals, $x^{-y} = y^{-x}$. Then $(-x)^{-y} = (-y)^{-x}$ since $x$ and $y$ have the same parity. (The number of factors of $-1$ we introduced to each side differs by $x-y$, which is even.) So solving the problem where $x$ and $y$ are negative reduces to solving it when $x$ and $y$ are positive.
answered Nov 19 '10 at 19:49
Michael LugoMichael Lugo
18.4k33576
$endgroup$
add a comment |
$begingroup$
Say $x^y = y^x$, and $x > y > 0$. Taking logs, $y log x = x log y$; rearranging, $(log x)/x = (log y)/y$. Let $f(x) = (log x)/x$; then this is $f(x) = f(y)$.
Now, $f^prime(x) = (1-log x)/x^2$, so $f$ is increasing for $x<e$ and decreasing for $x>e$. So if $x^y = y^x$ has a solution, then $x > e > y$. So $y$ must be $1$ or $2$. But $y = 1$ doesn't work. $y=2$ gives $x=4$.
(I've always thought of this as the ``standard'' solution to this problem and I'm a bit surprised nobody has posted it yet.)
If $x > 0 > y$, then $0 < x^y < 1$ and $y^x$ is an integer, so there are no such solutions.
If $0 > x > y$, then $x^y = y^x$ implies $x$ and $y$ must have the same parity. Also, taking reciprocals, $x^{-y} = y^{-x}$. Then $(-x)^{-y} = (-y)^{-x}$ since $x$ and $y$ have the same parity. (The number of factors of $-1$ we introduced to each side differs by $x-y$, which is even.) So solving the problem where $x$ and $y$ are negative reduces to solving it when $x$ and $y$ are positive.
answered Nov 19 '10 at 19:49
Michael LugoMichael Lugo
18.4k33576
$endgroup$
add a comment |
$begingroup$
Say $x^y = y^x$, and $x > y > 0$. Taking logs, $y log x = x log y$; rearranging, $(log x)/x = (log y)/y$. Let $f(x) = (log x)/x$; then this is $f(x) = f(y)$.
Now, $f^prime(x) = (1-log x)/x^2$, so $f$ is increasing for $x<e$ and decreasing for $x>e$. So if $x^y = y^x$ has a solution, then $x > e > y$. So $y$ must be $1$ or $2$. But $y = 1$ doesn't work. $y=2$ gives $x=4$.
(I've always thought of this as the ``standard'' solution to this problem and I'm a bit surprised nobody has posted it yet.)
If $x > 0 > y$, then $0 < x^y < 1$ and $y^x$ is an integer, so there are no such solutions.
If $0 > x > y$, then $x^y = y^x$ implies $x$ and $y$ must have the same parity. Also, taking reciprocals, $x^{-y} = y^{-x}$. Then $(-x)^{-y} = (-y)^{-x}$ since $x$ and $y$ have the same parity. (The number of factors of $-1$ we introduced to each side differs by $x-y$, which is even.) So solving the problem where $x$ and $y$ are negative reduces to solving it when $x$ and $y$ are positive.
answered Nov 19 '10 at 19:49
Michael LugoMichael Lugo
18.4k33576
$endgroup$
Say $x^y = y^x$, and $x > y > 0$. Taking logs, $y log x = x log y$; rearranging, $(log x)/x = (log y)/y$. Let $f(x) = (log x)/x$; then this is $f(x) = f(y)$.
Now, $f^prime(x) = (1-log x)/x^2$, so $f$ is increasing for $x<e$ and decreasing for $x>e$. So if $x^y = y^x$ has a solution, then $x > e > y$. So $y$ must be $1$ or $2$. But $y = 1$ doesn't work. $y=2$ gives $x=4$.
(I've always thought of this as the ``standard'' solution to this problem and I'm a bit surprised nobody has posted it yet.)
If $x > 0 > y$, then $0 < x^y < 1$ and $y^x$ is an integer, so there are no such solutions.
If $0 > x > y$, then $x^y = y^x$ implies $x$ and $y$ must have the same parity. Also, taking reciprocals, $x^{-y} = y^{-x}$. Then $(-x)^{-y} = (-y)^{-x}$ since $x$ and $y$ have the same parity. (The number of factors of $-1$ we introduced to each side differs by $x-y$, which is even.) So solving the problem where $x$ and $y$ are negative reduces to solving it when $x$ and $y$ are positive.
answered Nov 19 '10 at 19:49
Michael LugoMichael Lugo
18.4k33576
answered Nov 19 '10 at 19:49
Michael LugoMichael Lugo
18.4k33576
answered Nov 19 '10 at 19:49
Michael LugoMichael Lugo
18.4k33576
answered Nov 19 '10 at 19:49
Michael LugoMichael Lugo
18.4k33576
18.4k33576
add a comment |
add a comment |
$begingroup$
Although this thing has already been answered, here a shorter proof
Because $x^y = y^x $ is symmetric we first demand that $x>y$
Then we proceed simply this way:
$ x^y = y^x $
$ x = y^{frac x y } $
$ frac x y = y^{frac x y -1} $
$ frac x y -1 = y^{frac x y -1} - 1 $
Now we expand the rhs into its well-known exponential-series
$ frac x y -1 = ln(y)*(frac x y -1) + frac {((ln(y)*(frac x y -1))^2}{2!} + ... $
Here by the definition x>y the lhs is positive, so if $ ln(y) $ >=1 we had lhs $lt$ rhs
Thus $ ln(y) $ must be smaller than 1, and the only integer y>1 whose log is smaller than 1 is y=2, so there is the only possibility $y = 2$ and we are done.
[update] Well, after having determined $y=2$ the same procedure can be used to show, that after manyally checking $x=3$ (impossible) $x=4$ (possible) no $x>4$ can be chosen.
We ask for $x=4^{1+delta} ,delta > 0 $ inserting the value 2 for y:
$ 4^{(1+delta)*2}=2^{4^{(1+delta)}} $
Take log to base 2:
$ (1+delta)*4=4^{(1+delta)} $
$ delta =4^{delta} - 1 $
$ delta = ln(4)*delta + frac { (ln(4)*delta)^2 }{2!} + ldots $
$ 0 = (ln(4)-1)*delta + frac { (ln(4)*delta)^2 }{2!} + ldots $
Because $ ln(4)-1 >0 $ this can only be satisfied if $ delta =0 $
So indeed the only solutions, assuming x>y, is $ (x,y) = (4,2)$ .
[end update]
answered Nov 19 '10 at 17:02
Gottfried HelmsGottfried Helms
23.7k245101
$endgroup$
add a comment |
$begingroup$
Although this thing has already been answered, here a shorter proof
Because $x^y = y^x $ is symmetric we first demand that $x>y$
Then we proceed simply this way:
$ x^y = y^x $
$ x = y^{frac x y } $
$ frac x y = y^{frac x y -1} $
$ frac x y -1 = y^{frac x y -1} - 1 $
Now we expand the rhs into its well-known exponential-series
$ frac x y -1 = ln(y)*(frac x y -1) + frac {((ln(y)*(frac x y -1))^2}{2!} + ... $
Here by the definition x>y the lhs is positive, so if $ ln(y) $ >=1 we had lhs $lt$ rhs
Thus $ ln(y) $ must be smaller than 1, and the only integer y>1 whose log is smaller than 1 is y=2, so there is the only possibility $y = 2$ and we are done.
[update] Well, after having determined $y=2$ the same procedure can be used to show, that after manyally checking $x=3$ (impossible) $x=4$ (possible) no $x>4$ can be chosen.
We ask for $x=4^{1+delta} ,delta > 0 $ inserting the value 2 for y:
$ 4^{(1+delta)*2}=2^{4^{(1+delta)}} $
Take log to base 2:
$ (1+delta)*4=4^{(1+delta)} $
$ delta =4^{delta} - 1 $
$ delta = ln(4)*delta + frac { (ln(4)*delta)^2 }{2!} + ldots $
$ 0 = (ln(4)-1)*delta + frac { (ln(4)*delta)^2 }{2!} + ldots $
Because $ ln(4)-1 >0 $ this can only be satisfied if $ delta =0 $
So indeed the only solutions, assuming x>y, is $ (x,y) = (4,2)$ .
[end update]
answered Nov 19 '10 at 17:02
Gottfried HelmsGottfried Helms
23.7k245101
$endgroup$
add a comment |
$begingroup$
Although this thing has already been answered, here a shorter proof
Because $x^y = y^x $ is symmetric we first demand that $x>y$
Then we proceed simply this way:
$ x^y = y^x $
$ x = y^{frac x y } $
$ frac x y = y^{frac x y -1} $
$ frac x y -1 = y^{frac x y -1} - 1 $
Now we expand the rhs into its well-known exponential-series
$ frac x y -1 = ln(y)*(frac x y -1) + frac {((ln(y)*(frac x y -1))^2}{2!} + ... $
Here by the definition x>y the lhs is positive, so if $ ln(y) $ >=1 we had lhs $lt$ rhs
Thus $ ln(y) $ must be smaller than 1, and the only integer y>1 whose log is smaller than 1 is y=2, so there is the only possibility $y = 2$ and we are done.
[update] Well, after having determined $y=2$ the same procedure can be used to show, that after manyally checking $x=3$ (impossible) $x=4$ (possible) no $x>4$ can be chosen.
We ask for $x=4^{1+delta} ,delta > 0 $ inserting the value 2 for y:
$ 4^{(1+delta)*2}=2^{4^{(1+delta)}} $
Take log to base 2:
$ (1+delta)*4=4^{(1+delta)} $
$ delta =4^{delta} - 1 $
$ delta = ln(4)*delta + frac { (ln(4)*delta)^2 }{2!} + ldots $
$ 0 = (ln(4)-1)*delta + frac { (ln(4)*delta)^2 }{2!} + ldots $
Because $ ln(4)-1 >0 $ this can only be satisfied if $ delta =0 $
So indeed the only solutions, assuming x>y, is $ (x,y) = (4,2)$ .
[end update]
answered Nov 19 '10 at 17:02
Gottfried HelmsGottfried Helms
23.7k245101
$endgroup$
Although this thing has already been answered, here a shorter proof
Because $x^y = y^x $ is symmetric we first demand that $x>y$
Then we proceed simply this way:
$ x^y = y^x $
$ x = y^{frac x y } $
$ frac x y = y^{frac x y -1} $
$ frac x y -1 = y^{frac x y -1} - 1 $
Now we expand the rhs into its well-known exponential-series
$ frac x y -1 = ln(y)*(frac x y -1) + frac {((ln(y)*(frac x y -1))^2}{2!} + ... $
Here by the definition x>y the lhs is positive, so if $ ln(y) $ >=1 we had lhs $lt$ rhs
Thus $ ln(y) $ must be smaller than 1, and the only integer y>1 whose log is smaller than 1 is y=2, so there is the only possibility $y = 2$ and we are done.
[update] Well, after having determined $y=2$ the same procedure can be used to show, that after manyally checking $x=3$ (impossible) $x=4$ (possible) no $x>4$ can be chosen.
We ask for $x=4^{1+delta} ,delta > 0 $ inserting the value 2 for y:
$ 4^{(1+delta)*2}=2^{4^{(1+delta)}} $
Take log to base 2:
$ (1+delta)*4=4^{(1+delta)} $
$ delta =4^{delta} - 1 $
$ delta = ln(4)*delta + frac { (ln(4)*delta)^2 }{2!} + ldots $
$ 0 = (ln(4)-1)*delta + frac { (ln(4)*delta)^2 }{2!} + ldots $
Because $ ln(4)-1 >0 $ this can only be satisfied if $ delta =0 $
So indeed the only solutions, assuming x>y, is $ (x,y) = (4,2)$ .
[end update]
answered Nov 19 '10 at 17:02
Gottfried HelmsGottfried Helms
23.7k245101
edited Nov 19 '10 at 18:11
answered Nov 19 '10 at 17:02
Gottfried HelmsGottfried Helms
23.7k245101
answered Nov 19 '10 at 17:02
Gottfried HelmsGottfried Helms
23.7k245101
answered Nov 19 '10 at 17:02
Gottfried HelmsGottfried Helms
23.7k245101
23.7k245101
add a comment |
add a comment |
$begingroup$
Well I finally found an answer relating to some number theory I suppose !
Assume that : $x={p_1}^{alpha _ 1}.{p_2}^{alpha _ 2}...{p_k}^{alpha _ k}$ it is clear that number y prime factors are the same as number x but with different powers i.e: $y={p_1}^{beta _ 1}.{p_2}^{beta _ 2}...{p_k}^{beta _ k}$ replacing the first equation we get:
${({p_1}^{alpha _ 1}.{p_2}^{alpha _ 2}...{p_k}^{alpha _ k})}^y={({p_1}^{beta _ 1}.{p_2}^{beta _ 2}...{p_k}^{beta _ k})}^x$ i.e: ${p_1}^{{alpha _ 1}y}.{p_2}^{{alpha _ 2}y}...{p_k}^{{alpha _ k}y}={p_1}^{{beta _ 1}x}.{p_2}^{{beta _ 2}x}...{p_k}^{{beta _ k}x}$
Since the the powers ought to be equal we know for each $1le i le k$ we have:${alpha_i}y={beta_i}x$ i.e: ${alpha_i}/{beta_i}=x/y$
Considering that the equation is symmetric we can assume that $x le y$ but we have ${alpha_i}/{beta_i} = x/y ge 1$ hence ${alpha_i} ge {beta_i}$
Assume this obvious,easy-to-prove theorem:
Theorem #1
Consider $x,y in mathbb{N}$ such that $x={p_1}^{alpha _ 1}.{p_2}^{alpha _ 2}...{p_k}^{alpha _ k}$ $y={p_1}^{beta _ 1}.{p_2}^{beta _ 2}...{p_k}^{beta _ k}$ for each $1le i le k$ we have:
$y|x to {alpha_i}ge{beta_i}$ or vice versa
Using the Theorem #1 we can get that $y|x$ i.e $x=yt$ replacing in the main equation we get:
$x^y=y^x to ({yt})^y=y^{({yt})} to yt=y^t$
Now we must find the answers to the equation $yt=y^t$ for $t=1$ it is obvious that for every $y in mathbb{N}$ the equation is valid.so one answer is $x=y$
Yet again for $t=2$ we must have $2y=y^2$ i.e $y=2$ and we can conclude that $x=4$ (using the equation $x=yt$)so another answer is $x=4$ $land$ y=2$ (or vice versa)
We show that for $tge3$ the equation is not valid anymore.
If $tge3$ then $ygt2$ we prove that with these terms the inequality $y^t gt yt$ stands.
$y^t={(y-1+1)}^t={(y-1)}^t+...+binom{t}{2} {(y-1)}^2 + binom{t}{1}(y-1) +1 gt binom{t}{2} {(y-1)}^2 + t(y-1) +1$
But we have $y-1gt1$ so:
$y^t gt binom{t}{2} {(y-1)}^2 + t(y-1) +1= frac {t(t-1)}{2} -t +1 +yt= frac {(t-2)(t-1)}{2} + yt gt yt$
So it is proved that for $tge3$ is not valid anymore.$bullet$
P.S: The equation is solved for positive integers yet the solution for all the integers is quite the same!(took me an hour to write this all,hope you like my solution)
answered Jul 27 '14 at 6:33
FuriousMathematicianFuriousMathematician
320214
$endgroup$
add a comment |
$begingroup$
Well I finally found an answer relating to some number theory I suppose !
Assume that : $x={p_1}^{alpha _ 1}.{p_2}^{alpha _ 2}...{p_k}^{alpha _ k}$ it is clear that number y prime factors are the same as number x but with different powers i.e: $y={p_1}^{beta _ 1}.{p_2}^{beta _ 2}...{p_k}^{beta _ k}$ replacing the first equation we get:
${({p_1}^{alpha _ 1}.{p_2}^{alpha _ 2}...{p_k}^{alpha _ k})}^y={({p_1}^{beta _ 1}.{p_2}^{beta _ 2}...{p_k}^{beta _ k})}^x$ i.e: ${p_1}^{{alpha _ 1}y}.{p_2}^{{alpha _ 2}y}...{p_k}^{{alpha _ k}y}={p_1}^{{beta _ 1}x}.{p_2}^{{beta _ 2}x}...{p_k}^{{beta _ k}x}$
Since the the powers ought to be equal we know for each $1le i le k$ we have:${alpha_i}y={beta_i}x$ i.e: ${alpha_i}/{beta_i}=x/y$
Considering that the equation is symmetric we can assume that $x le y$ but we have ${alpha_i}/{beta_i} = x/y ge 1$ hence ${alpha_i} ge {beta_i}$
Assume this obvious,easy-to-prove theorem:
Theorem #1
Consider $x,y in mathbb{N}$ such that $x={p_1}^{alpha _ 1}.{p_2}^{alpha _ 2}...{p_k}^{alpha _ k}$ $y={p_1}^{beta _ 1}.{p_2}^{beta _ 2}...{p_k}^{beta _ k}$ for each $1le i le k$ we have:
$y|x to {alpha_i}ge{beta_i}$ or vice versa
Using the Theorem #1 we can get that $y|x$ i.e $x=yt$ replacing in the main equation we get:
$x^y=y^x to ({yt})^y=y^{({yt})} to yt=y^t$
Now we must find the answers to the equation $yt=y^t$ for $t=1$ it is obvious that for every $y in mathbb{N}$ the equation is valid.so one answer is $x=y$
Yet again for $t=2$ we must have $2y=y^2$ i.e $y=2$ and we can conclude that $x=4$ (using the equation $x=yt$)so another answer is $x=4$ $land$ y=2$ (or vice versa)
We show that for $tge3$ the equation is not valid anymore.
If $tge3$ then $ygt2$ we prove that with these terms the inequality $y^t gt yt$ stands.
$y^t={(y-1+1)}^t={(y-1)}^t+...+binom{t}{2} {(y-1)}^2 + binom{t}{1}(y-1) +1 gt binom{t}{2} {(y-1)}^2 + t(y-1) +1$
But we have $y-1gt1$ so:
$y^t gt binom{t}{2} {(y-1)}^2 + t(y-1) +1= frac {t(t-1)}{2} -t +1 +yt= frac {(t-2)(t-1)}{2} + yt gt yt$
So it is proved that for $tge3$ is not valid anymore.$bullet$
P.S: The equation is solved for positive integers yet the solution for all the integers is quite the same!(took me an hour to write this all,hope you like my solution)
answered Jul 27 '14 at 6:33
FuriousMathematicianFuriousMathematician
320214
$endgroup$
add a comment |
$begingroup$
Well I finally found an answer relating to some number theory I suppose !
Assume that : $x={p_1}^{alpha _ 1}.{p_2}^{alpha _ 2}...{p_k}^{alpha _ k}$ it is clear that number y prime factors are the same as number x but with different powers i.e: $y={p_1}^{beta _ 1}.{p_2}^{beta _ 2}...{p_k}^{beta _ k}$ replacing the first equation we get:
${({p_1}^{alpha _ 1}.{p_2}^{alpha _ 2}...{p_k}^{alpha _ k})}^y={({p_1}^{beta _ 1}.{p_2}^{beta _ 2}...{p_k}^{beta _ k})}^x$ i.e: ${p_1}^{{alpha _ 1}y}.{p_2}^{{alpha _ 2}y}...{p_k}^{{alpha _ k}y}={p_1}^{{beta _ 1}x}.{p_2}^{{beta _ 2}x}...{p_k}^{{beta _ k}x}$
Since the the powers ought to be equal we know for each $1le i le k$ we have:${alpha_i}y={beta_i}x$ i.e: ${alpha_i}/{beta_i}=x/y$
Considering that the equation is symmetric we can assume that $x le y$ but we have ${alpha_i}/{beta_i} = x/y ge 1$ hence ${alpha_i} ge {beta_i}$
Assume this obvious,easy-to-prove theorem:
Theorem #1
Consider $x,y in mathbb{N}$ such that $x={p_1}^{alpha _ 1}.{p_2}^{alpha _ 2}...{p_k}^{alpha _ k}$ $y={p_1}^{beta _ 1}.{p_2}^{beta _ 2}...{p_k}^{beta _ k}$ for each $1le i le k$ we have:
$y|x to {alpha_i}ge{beta_i}$ or vice versa
Using the Theorem #1 we can get that $y|x$ i.e $x=yt$ replacing in the main equation we get:
$x^y=y^x to ({yt})^y=y^{({yt})} to yt=y^t$
Now we must find the answers to the equation $yt=y^t$ for $t=1$ it is obvious that for every $y in mathbb{N}$ the equation is valid.so one answer is $x=y$
Yet again for $t=2$ we must have $2y=y^2$ i.e $y=2$ and we can conclude that $x=4$ (using the equation $x=yt$)so another answer is $x=4$ $land$ y=2$ (or vice versa)
We show that for $tge3$ the equation is not valid anymore.
If $tge3$ then $ygt2$ we prove that with these terms the inequality $y^t gt yt$ stands.
$y^t={(y-1+1)}^t={(y-1)}^t+...+binom{t}{2} {(y-1)}^2 + binom{t}{1}(y-1) +1 gt binom{t}{2} {(y-1)}^2 + t(y-1) +1$
But we have $y-1gt1$ so:
$y^t gt binom{t}{2} {(y-1)}^2 + t(y-1) +1= frac {t(t-1)}{2} -t +1 +yt= frac {(t-2)(t-1)}{2} + yt gt yt$
So it is proved that for $tge3$ is not valid anymore.$bullet$
P.S: The equation is solved for positive integers yet the solution for all the integers is quite the same!(took me an hour to write this all,hope you like my solution)
answered Jul 27 '14 at 6:33
FuriousMathematicianFuriousMathematician
320214
$endgroup$
Well I finally found an answer relating to some number theory I suppose !
Assume that : $x={p_1}^{alpha _ 1}.{p_2}^{alpha _ 2}...{p_k}^{alpha _ k}$ it is clear that number y prime factors are the same as number x but with different powers i.e: $y={p_1}^{beta _ 1}.{p_2}^{beta _ 2}...{p_k}^{beta _ k}$ replacing the first equation we get:
${({p_1}^{alpha _ 1}.{p_2}^{alpha _ 2}...{p_k}^{alpha _ k})}^y={({p_1}^{beta _ 1}.{p_2}^{beta _ 2}...{p_k}^{beta _ k})}^x$ i.e: ${p_1}^{{alpha _ 1}y}.{p_2}^{{alpha _ 2}y}...{p_k}^{{alpha _ k}y}={p_1}^{{beta _ 1}x}.{p_2}^{{beta _ 2}x}...{p_k}^{{beta _ k}x}$
Since the the powers ought to be equal we know for each $1le i le k$ we have:${alpha_i}y={beta_i}x$ i.e: ${alpha_i}/{beta_i}=x/y$
Considering that the equation is symmetric we can assume that $x le y$ but we have ${alpha_i}/{beta_i} = x/y ge 1$ hence ${alpha_i} ge {beta_i}$
Assume this obvious,easy-to-prove theorem:
Theorem #1
Consider $x,y in mathbb{N}$ such that $x={p_1}^{alpha _ 1}.{p_2}^{alpha _ 2}...{p_k}^{alpha _ k}$ $y={p_1}^{beta _ 1}.{p_2}^{beta _ 2}...{p_k}^{beta _ k}$ for each $1le i le k$ we have:
$y|x to {alpha_i}ge{beta_i}$ or vice versa
Using the Theorem #1 we can get that $y|x$ i.e $x=yt$ replacing in the main equation we get:
$x^y=y^x to ({yt})^y=y^{({yt})} to yt=y^t$
Now we must find the answers to the equation $yt=y^t$ for $t=1$ it is obvious that for every $y in mathbb{N}$ the equation is valid.so one answer is $x=y$
Yet again for $t=2$ we must have $2y=y^2$ i.e $y=2$ and we can conclude that $x=4$ (using the equation $x=yt$)so another answer is $x=4$ $land$ y=2$ (or vice versa)
We show that for $tge3$ the equation is not valid anymore.
If $tge3$ then $ygt2$ we prove that with these terms the inequality $y^t gt yt$ stands.
$y^t={(y-1+1)}^t={(y-1)}^t+...+binom{t}{2} {(y-1)}^2 + binom{t}{1}(y-1) +1 gt binom{t}{2} {(y-1)}^2 + t(y-1) +1$
But we have $y-1gt1$ so:
$y^t gt binom{t}{2} {(y-1)}^2 + t(y-1) +1= frac {t(t-1)}{2} -t +1 +yt= frac {(t-2)(t-1)}{2} + yt gt yt$
So it is proved that for $tge3$ is not valid anymore.$bullet$
P.S: The equation is solved for positive integers yet the solution for all the integers is quite the same!(took me an hour to write this all,hope you like my solution)
answered Jul 27 '14 at 6:33
FuriousMathematicianFuriousMathematician
320214
edited Oct 31 '14 at 11:05
answered Jul 27 '14 at 6:33
FuriousMathematicianFuriousMathematician
320214
answered Jul 27 '14 at 6:33
FuriousMathematicianFuriousMathematician
320214
answered Jul 27 '14 at 6:33
FuriousMathematicianFuriousMathematician
320214
320214
add a comment |
add a comment |
$begingroup$
I've collected some references, feel free to add more of them. (Some of them are taken from other answers. And, of course, some of them can contain further interesting references.)
Online:
Nick's Mathematical Puzzles - Solution to puzzle 48: Exponential equation (Wayback Machine)
On Torsten Sillke's page: http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/x%5Ey-x%5Ey (Wayback Machine)
Wikipedia: Equation $x^y=y^x$
Papers:
Michael A. Bennett and Bruce Reznick: Positive Rational Solutions to $x^y = y^{mx}$ : A Number-Theoretic Excursion, The American Mathematical Monthly , Vol. 111, No. 1 (Jan., 2004), pp. 13-21; available at jstor, arxiv or at author's homepage.
Marta Sved: On the Rational Solutions of $x^y = y^x$, Mathematics Magazine, Vol. 63, No. 1 (Feb., 1990), pp. 30-33, available at jstor. It is mentioned here, that this problem appeared in 1960 Putnam Competition (for integers)
F. Gerrish: 76.25 $a^{b}=b^{a}$: The Positive Integer Solution, The Mathematical Gazette, Vol. 76, No. 477 (Nov., 1992), p. 403. Jstor link
Solomon Hurwitz: On the Rational Solutions of $m^n=n^m$ with $mne n$, The American Mathematical Monthly, Vol. 74, No. 3 (Mar., 1967), pp. 298-300. jstor
Joel Anderson: Iterated Exponentials, The American Mathematical Monthly , Vol. 111, No. 8 (Oct., 2004), pp. 668-679. jstor
R. Arthur Knoebel, Exponentials Reiterated. The American Mathematical Monthly , Vol. 88, No. 4 (Apr., 1981), pp. 235-252 jstor, link
Books:
Andrew M. Gleason, R. E. Greenwood, Leroy Milton Kelly: William Lowell Putnam mathematical competition problems and solutions 1938-1964, ISBN 0883854287, p.59 and p.538
Titu Andreescu, Dorin Andrica, Ion Cucurezeanu: An Introduction to Diophantine Equations: A Problem-Based Approach, Springer, New York, 2010. Page 209
Searches:
The reason I've added this is that it can be somewhat tricky to search for a formula or an equation. So any interesting idea which could help finding interesting references may be of interest.
- diophantine "x^y=y^x"
$endgroup$
add a comment |
$begingroup$
I've collected some references, feel free to add more of them. (Some of them are taken from other answers. And, of course, some of them can contain further interesting references.)
Online:
Nick's Mathematical Puzzles - Solution to puzzle 48: Exponential equation (Wayback Machine)
On Torsten Sillke's page: http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/x%5Ey-x%5Ey (Wayback Machine)
Wikipedia: Equation $x^y=y^x$
Papers:
Michael A. Bennett and Bruce Reznick: Positive Rational Solutions to $x^y = y^{mx}$ : A Number-Theoretic Excursion, The American Mathematical Monthly , Vol. 111, No. 1 (Jan., 2004), pp. 13-21; available at jstor, arxiv or at author's homepage.
Marta Sved: On the Rational Solutions of $x^y = y^x$, Mathematics Magazine, Vol. 63, No. 1 (Feb., 1990), pp. 30-33, available at jstor. It is mentioned here, that this problem appeared in 1960 Putnam Competition (for integers)
F. Gerrish: 76.25 $a^{b}=b^{a}$: The Positive Integer Solution, The Mathematical Gazette, Vol. 76, No. 477 (Nov., 1992), p. 403. Jstor link
Solomon Hurwitz: On the Rational Solutions of $m^n=n^m$ with $mne n$, The American Mathematical Monthly, Vol. 74, No. 3 (Mar., 1967), pp. 298-300. jstor
Joel Anderson: Iterated Exponentials, The American Mathematical Monthly , Vol. 111, No. 8 (Oct., 2004), pp. 668-679. jstor
R. Arthur Knoebel, Exponentials Reiterated. The American Mathematical Monthly , Vol. 88, No. 4 (Apr., 1981), pp. 235-252 jstor, link
Books:
Andrew M. Gleason, R. E. Greenwood, Leroy Milton Kelly: William Lowell Putnam mathematical competition problems and solutions 1938-1964, ISBN 0883854287, p.59 and p.538
Titu Andreescu, Dorin Andrica, Ion Cucurezeanu: An Introduction to Diophantine Equations: A Problem-Based Approach, Springer, New York, 2010. Page 209
Searches:
The reason I've added this is that it can be somewhat tricky to search for a formula or an equation. So any interesting idea which could help finding interesting references may be of interest.
- diophantine "x^y=y^x"
$endgroup$
add a comment |
$begingroup$
I've collected some references, feel free to add more of them. (Some of them are taken from other answers. And, of course, some of them can contain further interesting references.)
Online:
Nick's Mathematical Puzzles - Solution to puzzle 48: Exponential equation (Wayback Machine)
On Torsten Sillke's page: http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/x%5Ey-x%5Ey (Wayback Machine)
Wikipedia: Equation $x^y=y^x$
Papers:
Michael A. Bennett and Bruce Reznick: Positive Rational Solutions to $x^y = y^{mx}$ : A Number-Theoretic Excursion, The American Mathematical Monthly , Vol. 111, No. 1 (Jan., 2004), pp. 13-21; available at jstor, arxiv or at author's homepage.
Marta Sved: On the Rational Solutions of $x^y = y^x$, Mathematics Magazine, Vol. 63, No. 1 (Feb., 1990), pp. 30-33, available at jstor. It is mentioned here, that this problem appeared in 1960 Putnam Competition (for integers)
F. Gerrish: 76.25 $a^{b}=b^{a}$: The Positive Integer Solution, The Mathematical Gazette, Vol. 76, No. 477 (Nov., 1992), p. 403. Jstor link
Solomon Hurwitz: On the Rational Solutions of $m^n=n^m$ with $mne n$, The American Mathematical Monthly, Vol. 74, No. 3 (Mar., 1967), pp. 298-300. jstor
Joel Anderson: Iterated Exponentials, The American Mathematical Monthly , Vol. 111, No. 8 (Oct., 2004), pp. 668-679. jstor
R. Arthur Knoebel, Exponentials Reiterated. The American Mathematical Monthly , Vol. 88, No. 4 (Apr., 1981), pp. 235-252 jstor, link
Books:
Andrew M. Gleason, R. E. Greenwood, Leroy Milton Kelly: William Lowell Putnam mathematical competition problems and solutions 1938-1964, ISBN 0883854287, p.59 and p.538
Titu Andreescu, Dorin Andrica, Ion Cucurezeanu: An Introduction to Diophantine Equations: A Problem-Based Approach, Springer, New York, 2010. Page 209
Searches:
The reason I've added this is that it can be somewhat tricky to search for a formula or an equation. So any interesting idea which could help finding interesting references may be of interest.
- diophantine "x^y=y^x"
$endgroup$
I've collected some references, feel free to add more of them. (Some of them are taken from other answers. And, of course, some of them can contain further interesting references.)
Online:
Nick's Mathematical Puzzles - Solution to puzzle 48: Exponential equation (Wayback Machine)
On Torsten Sillke's page: http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/x%5Ey-x%5Ey (Wayback Machine)
Wikipedia: Equation $x^y=y^x$
Papers:
Michael A. Bennett and Bruce Reznick: Positive Rational Solutions to $x^y = y^{mx}$ : A Number-Theoretic Excursion, The American Mathematical Monthly , Vol. 111, No. 1 (Jan., 2004), pp. 13-21; available at jstor, arxiv or at author's homepage.
Marta Sved: On the Rational Solutions of $x^y = y^x$, Mathematics Magazine, Vol. 63, No. 1 (Feb., 1990), pp. 30-33, available at jstor. It is mentioned here, that this problem appeared in 1960 Putnam Competition (for integers)
F. Gerrish: 76.25 $a^{b}=b^{a}$: The Positive Integer Solution, The Mathematical Gazette, Vol. 76, No. 477 (Nov., 1992), p. 403. Jstor link
Solomon Hurwitz: On the Rational Solutions of $m^n=n^m$ with $mne n$, The American Mathematical Monthly, Vol. 74, No. 3 (Mar., 1967), pp. 298-300. jstor
Joel Anderson: Iterated Exponentials, The American Mathematical Monthly , Vol. 111, No. 8 (Oct., 2004), pp. 668-679. jstor
R. Arthur Knoebel, Exponentials Reiterated. The American Mathematical Monthly , Vol. 88, No. 4 (Apr., 1981), pp. 235-252 jstor, link
Books:
Andrew M. Gleason, R. E. Greenwood, Leroy Milton Kelly: William Lowell Putnam mathematical competition problems and solutions 1938-1964, ISBN 0883854287, p.59 and p.538
Titu Andreescu, Dorin Andrica, Ion Cucurezeanu: An Introduction to Diophantine Equations: A Problem-Based Approach, Springer, New York, 2010. Page 209
Searches:
The reason I've added this is that it can be somewhat tricky to search for a formula or an equation. So any interesting idea which could help finding interesting references may be of interest.
- diophantine "x^y=y^x"
edited Feb 1 at 5:05
community wiki
7 revs
Martin Sleziak
add a comment |
add a comment |
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