Mixing
If a product of relatively prime integers is an $n$th power, then each is an $n$th power
$begingroup$
Show that if $a$, $b$, and $c$ are positive integers with $gcd(a, b) = 1$ and $ab = c^n$, then there are positive integers $d$, and $e$ such that $a = d^n$ and $b = e^n$.
elementary-number-theory
edited Feb 11 '11 at 19:45
Arturo Magidin
264k34588915
asked Feb 11 '11 at 14:55
kirakira
290114
$endgroup$
|
show 2 more comments
$begingroup$
Show that if $a$, $b$, and $c$ are positive integers with $gcd(a, b) = 1$ and $ab = c^n$, then there are positive integers $d$, and $e$ such that $a = d^n$ and $b = e^n$.
elementary-number-theory
edited Feb 11 '11 at 19:45
Arturo Magidin
264k34588915
asked Feb 11 '11 at 14:55
kirakira
290114
$endgroup$
3
$begingroup$
Have you considered the prime factorizations of $a$, $b$, and $c$?
$endgroup$
– Jon
Feb 11 '11 at 15:03
$begingroup$
1=ax+by => 1=gcd(a^n,b^n)
$endgroup$
– kira
Feb 11 '11 at 15:43
$begingroup$
@kira: Please ask questions, don't give orders. Also: please make your titles informative, not sentence fragments.
$endgroup$
– Arturo Magidin
Feb 11 '11 at 16:21
$begingroup$
obvious assuming the fundamental theorem of arithmetic
$endgroup$
– yoyo
Feb 11 '11 at 19:00
$begingroup$
@kira: The title describes your problem exactly; why did you change it back to not having mark-up and to the old phrasing?
$endgroup$
– Arturo Magidin
Feb 11 '11 at 19:46
|
show 2 more comments
$begingroup$
Show that if $a$, $b$, and $c$ are positive integers with $gcd(a, b) = 1$ and $ab = c^n$, then there are positive integers $d$, and $e$ such that $a = d^n$ and $b = e^n$.
elementary-number-theory
edited Feb 11 '11 at 19:45
Arturo Magidin
264k34588915
asked Feb 11 '11 at 14:55
kirakira
290114
$endgroup$
Show that if $a$, $b$, and $c$ are positive integers with $gcd(a, b) = 1$ and $ab = c^n$, then there are positive integers $d$, and $e$ such that $a = d^n$ and $b = e^n$.
elementary-number-theory
elementary-number-theory
edited Feb 11 '11 at 19:45
Arturo Magidin
264k34588915
asked Feb 11 '11 at 14:55
kirakira
290114
edited Feb 11 '11 at 19:45
Arturo Magidin
264k34588915
asked Feb 11 '11 at 14:55
kirakira
290114
edited Feb 11 '11 at 19:45
Arturo Magidin
264k34588915
edited Feb 11 '11 at 19:45
Arturo Magidin
264k34588915
edited Feb 11 '11 at 19:45
Arturo Magidin
264k34588915
264k34588915
asked Feb 11 '11 at 14:55
kirakira
290114
asked Feb 11 '11 at 14:55
kirakira
290114
asked Feb 11 '11 at 14:55
kirakira
290114
290114
3
$begingroup$
Have you considered the prime factorizations of $a$, $b$, and $c$?
$endgroup$
– Jon
Feb 11 '11 at 15:03
$begingroup$
1=ax+by => 1=gcd(a^n,b^n)
$endgroup$
– kira
Feb 11 '11 at 15:43
$begingroup$
@kira: Please ask questions, don't give orders. Also: please make your titles informative, not sentence fragments.
$endgroup$
– Arturo Magidin
Feb 11 '11 at 16:21
$begingroup$
obvious assuming the fundamental theorem of arithmetic
$endgroup$
– yoyo
Feb 11 '11 at 19:00
$begingroup$
@kira: The title describes your problem exactly; why did you change it back to not having mark-up and to the old phrasing?
$endgroup$
– Arturo Magidin
Feb 11 '11 at 19:46
|
show 2 more comments
3
$begingroup$
Have you considered the prime factorizations of $a$, $b$, and $c$?
$endgroup$
– Jon
Feb 11 '11 at 15:03
$begingroup$
1=ax+by => 1=gcd(a^n,b^n)
$endgroup$
– kira
Feb 11 '11 at 15:43
$begingroup$
@kira: Please ask questions, don't give orders. Also: please make your titles informative, not sentence fragments.
$endgroup$
– Arturo Magidin
Feb 11 '11 at 16:21
$begingroup$
obvious assuming the fundamental theorem of arithmetic
$endgroup$
– yoyo
Feb 11 '11 at 19:00
$begingroup$
@kira: The title describes your problem exactly; why did you change it back to not having mark-up and to the old phrasing?
$endgroup$
– Arturo Magidin
Feb 11 '11 at 19:46
3
3
$begingroup$
Have you considered the prime factorizations of $a$, $b$, and $c$?
$endgroup$
– Jon
Feb 11 '11 at 15:03
$begingroup$
Have you considered the prime factorizations of $a$, $b$, and $c$?
$endgroup$
– Jon
Feb 11 '11 at 15:03
$begingroup$
1=ax+by => 1=gcd(a^n,b^n)
$endgroup$
– kira
Feb 11 '11 at 15:43
$begingroup$
1=ax+by => 1=gcd(a^n,b^n)
$endgroup$
– kira
Feb 11 '11 at 15:43
$begingroup$
@kira: Please ask questions, don't give orders. Also: please make your titles informative, not sentence fragments.
$endgroup$
– Arturo Magidin
Feb 11 '11 at 16:21
$begingroup$
@kira: Please ask questions, don't give orders. Also: please make your titles informative, not sentence fragments.
$endgroup$
– Arturo Magidin
Feb 11 '11 at 16:21
$begingroup$
obvious assuming the fundamental theorem of arithmetic
$endgroup$
– yoyo
Feb 11 '11 at 19:00
$begingroup$
obvious assuming the fundamental theorem of arithmetic
$endgroup$
– yoyo
Feb 11 '11 at 19:00
$begingroup$
@kira: The title describes your problem exactly; why did you change it back to not having mark-up and to the old phrasing?
$endgroup$
– Arturo Magidin
Feb 11 '11 at 19:46
$begingroup$
@kira: The title describes your problem exactly; why did you change it back to not having mark-up and to the old phrasing?
$endgroup$
– Arturo Magidin
Feb 11 '11 at 19:46
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Of course it is easy using existence and uniqueness of prime factorizations. Below is a more general proof using gcd's (or ideals) that has the benefit of giving an explicit closed form.
$ab=c^n overset{rm Lemma}Rightarrow c=(a,c)(b,c) ,Rightarrow, ab = (a,c)^n(b,c)^nRightarrow dfrac{a}{(b,c)^n}! = dfrac{(a,c)^n}b$ $,Rightarrowbegin{align} a &= (a,c)^n\ b &= (b,c)^nend{align}$
where the last inference uses Unique Fractionization [both fractions are irreducible by $(a,b)!=!1$]
Lemma $ color{#c00}{cmid ab},, color{#0a0}{(a,b,c)=1} Rightarrow c = (a,c)(b,c) [=, (ab,ccolor{#0a0}{(a,b,c)}) = (color{#c00}{ab,c}) = c,]$
Remark $ $ Alternatively $ (a,c)^n! = (a^n,c^n) = (a^n,ab) = a(a^n,b) = a$ and $,(b,c)^n = b,$ by symmetry where the first equality employs the Freshman's Dream.
As $ $ Weil remarks, $ $ this result can be viewed as the essence of Fermat's method of infinite descent. $ $ It generalizes to rings of algebraic integers but depends upon much deeper results in this more general context, viz. the finiteness of the class number and Dirichlet's unit theorem.
answered Feb 12 '11 at 3:30
Bill DubuqueBill Dubuque
211k29194648
$endgroup$
add a comment |
$begingroup$
Been a while since I did any maths, so this is probably the wrong way of going about it. I'd take logs with base $a$ of $ab=c^n$, this will give:
$$1 + log_a(b) = nlog_a(c) rightarrow 1 = nlog_a(c) - log_a(b).$$
For the right hand side to equal 1, we need $b = e^n$ (this wouldn't necessarily be the case if $gcd(a,b) neq 1$):
$$1 = n(log_a(c) - log_a(e)).$$
I'd take the inverse log from here, giving $a = left(frac{c}{e}right)^n$. Simple to explain why $frac{c}{e}$ must be a whole number, $d$.
Bet that's the worst possible solution to this problem
edited Feb 12 '11 at 6:00
Arturo Magidin
264k34588915
answered Feb 11 '11 at 15:45
user6915user6915
1191
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Of course it is easy using existence and uniqueness of prime factorizations. Below is a more general proof using gcd's (or ideals) that has the benefit of giving an explicit closed form.
$ab=c^n overset{rm Lemma}Rightarrow c=(a,c)(b,c) ,Rightarrow, ab = (a,c)^n(b,c)^nRightarrow dfrac{a}{(b,c)^n}! = dfrac{(a,c)^n}b$ $,Rightarrowbegin{align} a &= (a,c)^n\ b &= (b,c)^nend{align}$
where the last inference uses Unique Fractionization [both fractions are irreducible by $(a,b)!=!1$]
Lemma $ color{#c00}{cmid ab},, color{#0a0}{(a,b,c)=1} Rightarrow c = (a,c)(b,c) [=, (ab,ccolor{#0a0}{(a,b,c)}) = (color{#c00}{ab,c}) = c,]$
Remark $ $ Alternatively $ (a,c)^n! = (a^n,c^n) = (a^n,ab) = a(a^n,b) = a$ and $,(b,c)^n = b,$ by symmetry where the first equality employs the Freshman's Dream.
As $ $ Weil remarks, $ $ this result can be viewed as the essence of Fermat's method of infinite descent. $ $ It generalizes to rings of algebraic integers but depends upon much deeper results in this more general context, viz. the finiteness of the class number and Dirichlet's unit theorem.
answered Feb 12 '11 at 3:30
Bill DubuqueBill Dubuque
211k29194648
$endgroup$
add a comment |
$begingroup$
Of course it is easy using existence and uniqueness of prime factorizations. Below is a more general proof using gcd's (or ideals) that has the benefit of giving an explicit closed form.
$ab=c^n overset{rm Lemma}Rightarrow c=(a,c)(b,c) ,Rightarrow, ab = (a,c)^n(b,c)^nRightarrow dfrac{a}{(b,c)^n}! = dfrac{(a,c)^n}b$ $,Rightarrowbegin{align} a &= (a,c)^n\ b &= (b,c)^nend{align}$
where the last inference uses Unique Fractionization [both fractions are irreducible by $(a,b)!=!1$]
Lemma $ color{#c00}{cmid ab},, color{#0a0}{(a,b,c)=1} Rightarrow c = (a,c)(b,c) [=, (ab,ccolor{#0a0}{(a,b,c)}) = (color{#c00}{ab,c}) = c,]$
Remark $ $ Alternatively $ (a,c)^n! = (a^n,c^n) = (a^n,ab) = a(a^n,b) = a$ and $,(b,c)^n = b,$ by symmetry where the first equality employs the Freshman's Dream.
As $ $ Weil remarks, $ $ this result can be viewed as the essence of Fermat's method of infinite descent. $ $ It generalizes to rings of algebraic integers but depends upon much deeper results in this more general context, viz. the finiteness of the class number and Dirichlet's unit theorem.
answered Feb 12 '11 at 3:30
Bill DubuqueBill Dubuque
211k29194648
$endgroup$
add a comment |
$begingroup$
Of course it is easy using existence and uniqueness of prime factorizations. Below is a more general proof using gcd's (or ideals) that has the benefit of giving an explicit closed form.
$ab=c^n overset{rm Lemma}Rightarrow c=(a,c)(b,c) ,Rightarrow, ab = (a,c)^n(b,c)^nRightarrow dfrac{a}{(b,c)^n}! = dfrac{(a,c)^n}b$ $,Rightarrowbegin{align} a &= (a,c)^n\ b &= (b,c)^nend{align}$
where the last inference uses Unique Fractionization [both fractions are irreducible by $(a,b)!=!1$]
Lemma $ color{#c00}{cmid ab},, color{#0a0}{(a,b,c)=1} Rightarrow c = (a,c)(b,c) [=, (ab,ccolor{#0a0}{(a,b,c)}) = (color{#c00}{ab,c}) = c,]$
Remark $ $ Alternatively $ (a,c)^n! = (a^n,c^n) = (a^n,ab) = a(a^n,b) = a$ and $,(b,c)^n = b,$ by symmetry where the first equality employs the Freshman's Dream.
As $ $ Weil remarks, $ $ this result can be viewed as the essence of Fermat's method of infinite descent. $ $ It generalizes to rings of algebraic integers but depends upon much deeper results in this more general context, viz. the finiteness of the class number and Dirichlet's unit theorem.
answered Feb 12 '11 at 3:30
Bill DubuqueBill Dubuque
211k29194648
$endgroup$
Of course it is easy using existence and uniqueness of prime factorizations. Below is a more general proof using gcd's (or ideals) that has the benefit of giving an explicit closed form.
$ab=c^n overset{rm Lemma}Rightarrow c=(a,c)(b,c) ,Rightarrow, ab = (a,c)^n(b,c)^nRightarrow dfrac{a}{(b,c)^n}! = dfrac{(a,c)^n}b$ $,Rightarrowbegin{align} a &= (a,c)^n\ b &= (b,c)^nend{align}$
where the last inference uses Unique Fractionization [both fractions are irreducible by $(a,b)!=!1$]
Lemma $ color{#c00}{cmid ab},, color{#0a0}{(a,b,c)=1} Rightarrow c = (a,c)(b,c) [=, (ab,ccolor{#0a0}{(a,b,c)}) = (color{#c00}{ab,c}) = c,]$
Remark $ $ Alternatively $ (a,c)^n! = (a^n,c^n) = (a^n,ab) = a(a^n,b) = a$ and $,(b,c)^n = b,$ by symmetry where the first equality employs the Freshman's Dream.
As $ $ Weil remarks, $ $ this result can be viewed as the essence of Fermat's method of infinite descent. $ $ It generalizes to rings of algebraic integers but depends upon much deeper results in this more general context, viz. the finiteness of the class number and Dirichlet's unit theorem.
answered Feb 12 '11 at 3:30
Bill DubuqueBill Dubuque
211k29194648
edited Feb 24 at 16:29
answered Feb 12 '11 at 3:30
Bill DubuqueBill Dubuque
211k29194648
answered Feb 12 '11 at 3:30
Bill DubuqueBill Dubuque
211k29194648
answered Feb 12 '11 at 3:30
Bill DubuqueBill Dubuque
211k29194648
211k29194648
add a comment |
add a comment |
$begingroup$
Been a while since I did any maths, so this is probably the wrong way of going about it. I'd take logs with base $a$ of $ab=c^n$, this will give:
$$1 + log_a(b) = nlog_a(c) rightarrow 1 = nlog_a(c) - log_a(b).$$
For the right hand side to equal 1, we need $b = e^n$ (this wouldn't necessarily be the case if $gcd(a,b) neq 1$):
$$1 = n(log_a(c) - log_a(e)).$$
I'd take the inverse log from here, giving $a = left(frac{c}{e}right)^n$. Simple to explain why $frac{c}{e}$ must be a whole number, $d$.
Bet that's the worst possible solution to this problem
edited Feb 12 '11 at 6:00
Arturo Magidin
264k34588915
answered Feb 11 '11 at 15:45
user6915user6915
1191
$endgroup$
add a comment |
$begingroup$
Been a while since I did any maths, so this is probably the wrong way of going about it. I'd take logs with base $a$ of $ab=c^n$, this will give:
$$1 + log_a(b) = nlog_a(c) rightarrow 1 = nlog_a(c) - log_a(b).$$
For the right hand side to equal 1, we need $b = e^n$ (this wouldn't necessarily be the case if $gcd(a,b) neq 1$):
$$1 = n(log_a(c) - log_a(e)).$$
I'd take the inverse log from here, giving $a = left(frac{c}{e}right)^n$. Simple to explain why $frac{c}{e}$ must be a whole number, $d$.
Bet that's the worst possible solution to this problem
edited Feb 12 '11 at 6:00
Arturo Magidin
264k34588915
answered Feb 11 '11 at 15:45
user6915user6915
1191
$endgroup$
add a comment |
$begingroup$
Been a while since I did any maths, so this is probably the wrong way of going about it. I'd take logs with base $a$ of $ab=c^n$, this will give:
$$1 + log_a(b) = nlog_a(c) rightarrow 1 = nlog_a(c) - log_a(b).$$
For the right hand side to equal 1, we need $b = e^n$ (this wouldn't necessarily be the case if $gcd(a,b) neq 1$):
$$1 = n(log_a(c) - log_a(e)).$$
I'd take the inverse log from here, giving $a = left(frac{c}{e}right)^n$. Simple to explain why $frac{c}{e}$ must be a whole number, $d$.
Bet that's the worst possible solution to this problem
edited Feb 12 '11 at 6:00
Arturo Magidin
264k34588915
answered Feb 11 '11 at 15:45
user6915user6915
1191
$endgroup$
Been a while since I did any maths, so this is probably the wrong way of going about it. I'd take logs with base $a$ of $ab=c^n$, this will give:
$$1 + log_a(b) = nlog_a(c) rightarrow 1 = nlog_a(c) - log_a(b).$$
For the right hand side to equal 1, we need $b = e^n$ (this wouldn't necessarily be the case if $gcd(a,b) neq 1$):
$$1 = n(log_a(c) - log_a(e)).$$
I'd take the inverse log from here, giving $a = left(frac{c}{e}right)^n$. Simple to explain why $frac{c}{e}$ must be a whole number, $d$.
Bet that's the worst possible solution to this problem
edited Feb 12 '11 at 6:00
Arturo Magidin
264k34588915
answered Feb 11 '11 at 15:45
user6915user6915
1191
edited Feb 12 '11 at 6:00
Arturo Magidin
264k34588915
edited Feb 12 '11 at 6:00
Arturo Magidin
264k34588915
edited Feb 12 '11 at 6:00
Arturo Magidin
264k34588915
264k34588915
answered Feb 11 '11 at 15:45
user6915user6915
1191
answered Feb 11 '11 at 15:45
user6915user6915
1191
answered Feb 11 '11 at 15:45
user6915user6915
1191
1191
add a comment |
add a comment |
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3
$begingroup$
Have you considered the prime factorizations of $a$, $b$, and $c$?
$endgroup$
– Jon
Feb 11 '11 at 15:03
$begingroup$
1=ax+by => 1=gcd(a^n,b^n)
$endgroup$
– kira
Feb 11 '11 at 15:43
$begingroup$
@kira: Please ask questions, don't give orders. Also: please make your titles informative, not sentence fragments.
$endgroup$
– Arturo Magidin
Feb 11 '11 at 16:21
$begingroup$
obvious assuming the fundamental theorem of arithmetic
$endgroup$
– yoyo
Feb 11 '11 at 19:00
$begingroup$
@kira: The title describes your problem exactly; why did you change it back to not having mark-up and to the old phrasing?
$endgroup$
– Arturo Magidin
Feb 11 '11 at 19:46