Mixing
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Show that if $a,b,cin mathbb{N}$ and ${a^2+b^2+c^2}over{abc+1}$ is an integer it is the sum of two nonzero...
$begingroup$
I was reading about the fascinating problem in the IMO ($1988 $ #$6$) that asks:
Let $a$ and $b$ be positive integers such that $(ab+1) | (a^2+b^2)$. Show that ${a^2+b^2}over{ab+1}$ is a perfect square.
I wondered what would happen with the natural extension of this problem to three constants. Upon exploration with Mathematica, I found that the integral values of ${a^2+b^2+c^2}over{abc+1}$ all seemed to be expressible as the sum of two nonzero perfect squares. The converse also seems to be true (given that a number is the sum of two nonzero squares, it seems to be expressible in the form of this fraction).
So I pose two questions:
If $a,b,cin mathbb{N}$ and $k={{a^2+b^2+c^2}over{abc+1}}$ is an integer, is $k$ the sum of two nonzero squares?
and
Given that $k$ is expressible as the sum of two nonzero perfect squares, do there exist $a,b,cinmathbb{N}$ for which ${{a^2+b^2+c^2}over{abc+1}}=k$?
number-theory square-numbers sums-of-squares vieta-jumping
edited Jan 31 at 2:54
Will Jagy
104k5102201
asked Jan 31 at 2:35
volcanrbvolcanrb
682216
$endgroup$
add a comment |
$begingroup$
I was reading about the fascinating problem in the IMO ($1988 $ #$6$) that asks:
Let $a$ and $b$ be positive integers such that $(ab+1) | (a^2+b^2)$. Show that ${a^2+b^2}over{ab+1}$ is a perfect square.
I wondered what would happen with the natural extension of this problem to three constants. Upon exploration with Mathematica, I found that the integral values of ${a^2+b^2+c^2}over{abc+1}$ all seemed to be expressible as the sum of two nonzero perfect squares. The converse also seems to be true (given that a number is the sum of two nonzero squares, it seems to be expressible in the form of this fraction).
So I pose two questions:
If $a,b,cin mathbb{N}$ and $k={{a^2+b^2+c^2}over{abc+1}}$ is an integer, is $k$ the sum of two nonzero squares?
and
Given that $k$ is expressible as the sum of two nonzero perfect squares, do there exist $a,b,cinmathbb{N}$ for which ${{a^2+b^2+c^2}over{abc+1}}=k$?
number-theory square-numbers sums-of-squares vieta-jumping
edited Jan 31 at 2:54
Will Jagy
104k5102201
asked Jan 31 at 2:35
volcanrbvolcanrb
682216
$endgroup$
$begingroup$
Does the usual method for the $(a^2+b^2)(ab+1)$ problem also work here?
$endgroup$
– Lord Shark the Unknown
Jan 31 at 2:48
$begingroup$
I unfortunately don't have enough experience to be able to solve that
$endgroup$
– volcanrb
Jan 31 at 2:50
1
$begingroup$
These days, this problem can be googled as well as solved: Searching for "IMO Problem 6" leads to en.wikipedia.org/wiki/Vieta_jumping
$endgroup$
– Lord Shark the Unknown
Jan 31 at 2:59
1
$begingroup$
To whoever downvoted: how can I improve the question?
$endgroup$
– volcanrb
Jan 31 at 3:11
add a comment |
$begingroup$
I was reading about the fascinating problem in the IMO ($1988 $ #$6$) that asks:
Let $a$ and $b$ be positive integers such that $(ab+1) | (a^2+b^2)$. Show that ${a^2+b^2}over{ab+1}$ is a perfect square.
I wondered what would happen with the natural extension of this problem to three constants. Upon exploration with Mathematica, I found that the integral values of ${a^2+b^2+c^2}over{abc+1}$ all seemed to be expressible as the sum of two nonzero perfect squares. The converse also seems to be true (given that a number is the sum of two nonzero squares, it seems to be expressible in the form of this fraction).
So I pose two questions:
If $a,b,cin mathbb{N}$ and $k={{a^2+b^2+c^2}over{abc+1}}$ is an integer, is $k$ the sum of two nonzero squares?
and
Given that $k$ is expressible as the sum of two nonzero perfect squares, do there exist $a,b,cinmathbb{N}$ for which ${{a^2+b^2+c^2}over{abc+1}}=k$?
number-theory square-numbers sums-of-squares vieta-jumping
edited Jan 31 at 2:54
Will Jagy
104k5102201
asked Jan 31 at 2:35
volcanrbvolcanrb
682216
$endgroup$
I was reading about the fascinating problem in the IMO ($1988 $ #$6$) that asks:
Let $a$ and $b$ be positive integers such that $(ab+1) | (a^2+b^2)$. Show that ${a^2+b^2}over{ab+1}$ is a perfect square.
I wondered what would happen with the natural extension of this problem to three constants. Upon exploration with Mathematica, I found that the integral values of ${a^2+b^2+c^2}over{abc+1}$ all seemed to be expressible as the sum of two nonzero perfect squares. The converse also seems to be true (given that a number is the sum of two nonzero squares, it seems to be expressible in the form of this fraction).
So I pose two questions:
If $a,b,cin mathbb{N}$ and $k={{a^2+b^2+c^2}over{abc+1}}$ is an integer, is $k$ the sum of two nonzero squares?
and
Given that $k$ is expressible as the sum of two nonzero perfect squares, do there exist $a,b,cinmathbb{N}$ for which ${{a^2+b^2+c^2}over{abc+1}}=k$?
number-theory square-numbers sums-of-squares vieta-jumping
number-theory square-numbers sums-of-squares vieta-jumping
edited Jan 31 at 2:54
Will Jagy
104k5102201
asked Jan 31 at 2:35
volcanrbvolcanrb
682216
edited Jan 31 at 2:54
Will Jagy
104k5102201
asked Jan 31 at 2:35
volcanrbvolcanrb
682216
edited Jan 31 at 2:54
Will Jagy
104k5102201
edited Jan 31 at 2:54
Will Jagy
104k5102201
edited Jan 31 at 2:54
Will Jagy
104k5102201
104k5102201
asked Jan 31 at 2:35
volcanrbvolcanrb
682216
asked Jan 31 at 2:35
volcanrbvolcanrb
682216
asked Jan 31 at 2:35
volcanrbvolcanrb
682216
682216
$begingroup$
Does the usual method for the $(a^2+b^2)(ab+1)$ problem also work here?
$endgroup$
– Lord Shark the Unknown
Jan 31 at 2:48
$begingroup$
I unfortunately don't have enough experience to be able to solve that
$endgroup$
– volcanrb
Jan 31 at 2:50
1
$begingroup$
These days, this problem can be googled as well as solved: Searching for "IMO Problem 6" leads to en.wikipedia.org/wiki/Vieta_jumping
$endgroup$
– Lord Shark the Unknown
Jan 31 at 2:59
1
$begingroup$
To whoever downvoted: how can I improve the question?
$endgroup$
– volcanrb
Jan 31 at 3:11
add a comment |
$begingroup$
Does the usual method for the $(a^2+b^2)(ab+1)$ problem also work here?
$endgroup$
– Lord Shark the Unknown
Jan 31 at 2:48
$begingroup$
I unfortunately don't have enough experience to be able to solve that
$endgroup$
– volcanrb
Jan 31 at 2:50
1
$begingroup$
These days, this problem can be googled as well as solved: Searching for "IMO Problem 6" leads to en.wikipedia.org/wiki/Vieta_jumping
$endgroup$
– Lord Shark the Unknown
Jan 31 at 2:59
1
$begingroup$
To whoever downvoted: how can I improve the question?
$endgroup$
– volcanrb
Jan 31 at 3:11
$begingroup$
Does the usual method for the $(a^2+b^2)(ab+1)$ problem also work here?
$endgroup$
– Lord Shark the Unknown
Jan 31 at 2:48
$begingroup$
Does the usual method for the $(a^2+b^2)(ab+1)$ problem also work here?
$endgroup$
– Lord Shark the Unknown
Jan 31 at 2:48
$begingroup$
I unfortunately don't have enough experience to be able to solve that
$endgroup$
– volcanrb
Jan 31 at 2:50
$begingroup$
I unfortunately don't have enough experience to be able to solve that
$endgroup$
– volcanrb
Jan 31 at 2:50
1
1
$begingroup$
These days, this problem can be googled as well as solved: Searching for "IMO Problem 6" leads to en.wikipedia.org/wiki/Vieta_jumping
$endgroup$
– Lord Shark the Unknown
Jan 31 at 2:59
$begingroup$
These days, this problem can be googled as well as solved: Searching for "IMO Problem 6" leads to en.wikipedia.org/wiki/Vieta_jumping
$endgroup$
– Lord Shark the Unknown
Jan 31 at 2:59
1
1
$begingroup$
To whoever downvoted: how can I improve the question?
$endgroup$
– volcanrb
Jan 31 at 3:11
$begingroup$
To whoever downvoted: how can I improve the question?
$endgroup$
– volcanrb
Jan 31 at 3:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I PROVED THE CONJECTURE
$$ x^2 + y^2 + z^2 = k + kxyz $$
Notice that we cannot have a solution with $x<0$
Got the other part. Let $x geq y geq z geq 1$ be a solution giving $k$ that minimizes $x+y+z$ among positive solutions giving $k.$ Vieta jumping is just the new solution
$$ (x,y,z) mapsto (kyz-x,y,z) $$
The equation is
$$ x^2 - (kyz)x + y^2 + z^2 - k = 0. $$
The replacement root $x'$ gives $x + x' = kyz. $ The product is $xx' = y^2 + z^2 - k.$
We have minimized $x+y+z$ among positive triples giving $k.$ We get a new triple with $x' = kyz - x.$ As it is impossible to have $x' <0,$ we have an Alternative:
(A) $kyz - x = 0$
(B) $kyz - x geq x$
We will show that alternative (B) does not occur for, say, $k > 2.$
ASSUME (B), namely $kyz geq 2x, $ with $x geq y geq z geq 1.$
$$ x^2 + y^2 + z^2 = k + (kyz)x geq k + (2x)x= k + 2x^2 $$
$$ color{red}{ y^2 + z^2 geq k + x^2}. $$
But
$$ y^2 - (kx)yz + z^2 = k - x^2. $$
Add,
$$ 2y^2 - (kx)yz + 2z^2 geq 2k . $$
$$ y^2 - left(frac{kx}{2} right)yz + z^2 geq k . $$
So, $y geq z geq 1$ and
$$ color{red}{y^2 - left(frac{kx}{2} right)yz + z^2 > 0 }. $$
From the quadratic formula,
$$ y > frac{1}{4} left( kx + sqrt {k^2 x^2 - 16} right) z . $$
When $k geq 4,$ we see $y > x,$ a contradiction. When $k=3$ and $z geq 2,$ we still get $y > x.$ Finally, with $k=3$ and $z=1,$ the condition $y^2 + z^2 geq k + x^2$ reads $y^2 + 1 geq 3 + x^2$ or $y^2 geq 2 + x^2,$ so again $y > x.$
These contradictions tell us that alternative (A) actually holds for $k geq 3.$ That is, $kyz = x,$ so $kyz - x = 0,$ and we have a new solution $(0,y,z)$ with $y geq z geq 1$ and this $x=0$ to
$$ x^2 + y^2 + z^2 = k + kxyz. $$ As this $x=0,$ we have reached
$$ color{red}{ y^2 + z^2 = k } $$
answered Jan 31 at 21:16
Will JagyWill Jagy
104k5102201
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add a comment |
$begingroup$
The method is called Vieta Jumping. It is a special case of the automorphism group of an indefinite binary quadratic form.
Part Two
The first part of your question: given $k = a^2 + b^2,$ you get an answer with
$$ x=a, ; ; y =b, ; ; z = kab $$
Then
$$ frac{x^2 + y^2 + z^2}{1+xyz} = k $$
Part One
The description of Vieta Jumping in links is not quite specific enough to settle this. The conjecture seems very likely; here are the values of $k$ with $ 5000 geq a geq b geq c geq 1,$ but ignoring the many, many ways to get $k=2.$
k a b c k
5 10 2 1 5 = 5
5 1102 230 1 5 = 5
5 230 48 1 5 = 5
5 2399 48 10 5 = 5
5 48 10 1 5 = 5
5 4948 99 10 5 = 5
5 980 99 2 5 = 5
5 99 10 2 5 = 5
8 32 2 2 8 = 2^3
8 510 32 2 8 = 2^3
10 2940 297 1 10 = 2 5
10 297 30 1 10 = 2 5
10 30 3 1 10 = 2 5
10 899 30 3 10 = 2 5
13 2025 78 2 13 = 13
13 3040 78 3 13 = 13
13 78 3 2 13 = 13
17 1152 68 1 17 = 17
17 4623 68 4 17 = 17
17 68 4 1 17 = 17
18 162 3 3 18 = 2 3^2
20 160 4 2 20 = 2^2 5
25 300 4 3 25 = 5^2
26 130 5 1 26 = 2 13
26 3375 130 1 26 = 2 13
29 290 5 2 29 = 29
32 512 4 4 32 = 2^5
34 510 5 3 34 = 2 17
37 222 6 1 37 = 37
40 480 6 2 40 = 2^3 5
41 820 5 4 41 = 41
45 810 6 3 45 = 3^2 5
50 1250 5 5 50 = 2 5^2
50 350 7 1 50 = 2 5^2
52 1248 6 4 52 = 2^2 13
53 742 7 2 53 = 53
58 1218 7 3 58 = 2 29
61 1830 6 5 61 = 61
65 1820 7 4 65 = 5 13
65 520 8 1 65 = 5 13
68 1088 8 2 68 = 2^2 17
72 2592 6 6 72 = 2^3 3^2
73 1752 8 3 73 = 73
74 2590 7 5 74 = 2 37
80 2560 8 4 80 = 2^4 5
82 738 9 1 82 = 2 41
85 1530 9 2 85 = 5 17
85 3570 7 6 85 = 5 17
89 3560 8 5 89 = 89
90 2430 9 3 90 = 2 3^2 5
97 3492 9 4 97 = 97
98 4802 7 7 98 = 2 7^2
100 4800 8 6 100 = 2^2 5^2
101 1010 10 1 101 = 101
104 2080 10 2 104 = 2^3 13
106 4770 9 5 106 = 2 53
109 3270 10 3 109 = 109
116 4640 10 4 116 = 2^2 29
122 1342 11 1 122 = 2 61
125 2750 11 2 125 = 5^3
130 4290 11 3 130 = 2 5 13
145 1740 12 1 145 = 5 29
148 3552 12 2 148 = 2^2 37
170 2210 13 1 170 = 2 5 17
173 4498 13 2 173 = 173
197 2758 14 1 197 = 197
226 3390 15 1 226 = 2 113
257 4112 16 1 257 = 257
290 4930 17 1 290 = 2 5 29
k a b c k
answered Jan 31 at 2:54
Will JagyWill Jagy
104k5102201
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I PROVED THE CONJECTURE
$$ x^2 + y^2 + z^2 = k + kxyz $$
Notice that we cannot have a solution with $x<0$
Got the other part. Let $x geq y geq z geq 1$ be a solution giving $k$ that minimizes $x+y+z$ among positive solutions giving $k.$ Vieta jumping is just the new solution
$$ (x,y,z) mapsto (kyz-x,y,z) $$
The equation is
$$ x^2 - (kyz)x + y^2 + z^2 - k = 0. $$
The replacement root $x'$ gives $x + x' = kyz. $ The product is $xx' = y^2 + z^2 - k.$
We have minimized $x+y+z$ among positive triples giving $k.$ We get a new triple with $x' = kyz - x.$ As it is impossible to have $x' <0,$ we have an Alternative:
(A) $kyz - x = 0$
(B) $kyz - x geq x$
We will show that alternative (B) does not occur for, say, $k > 2.$
ASSUME (B), namely $kyz geq 2x, $ with $x geq y geq z geq 1.$
$$ x^2 + y^2 + z^2 = k + (kyz)x geq k + (2x)x= k + 2x^2 $$
$$ color{red}{ y^2 + z^2 geq k + x^2}. $$
But
$$ y^2 - (kx)yz + z^2 = k - x^2. $$
Add,
$$ 2y^2 - (kx)yz + 2z^2 geq 2k . $$
$$ y^2 - left(frac{kx}{2} right)yz + z^2 geq k . $$
So, $y geq z geq 1$ and
$$ color{red}{y^2 - left(frac{kx}{2} right)yz + z^2 > 0 }. $$
From the quadratic formula,
$$ y > frac{1}{4} left( kx + sqrt {k^2 x^2 - 16} right) z . $$
When $k geq 4,$ we see $y > x,$ a contradiction. When $k=3$ and $z geq 2,$ we still get $y > x.$ Finally, with $k=3$ and $z=1,$ the condition $y^2 + z^2 geq k + x^2$ reads $y^2 + 1 geq 3 + x^2$ or $y^2 geq 2 + x^2,$ so again $y > x.$
These contradictions tell us that alternative (A) actually holds for $k geq 3.$ That is, $kyz = x,$ so $kyz - x = 0,$ and we have a new solution $(0,y,z)$ with $y geq z geq 1$ and this $x=0$ to
$$ x^2 + y^2 + z^2 = k + kxyz. $$ As this $x=0,$ we have reached
$$ color{red}{ y^2 + z^2 = k } $$
answered Jan 31 at 21:16
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
I PROVED THE CONJECTURE
$$ x^2 + y^2 + z^2 = k + kxyz $$
Notice that we cannot have a solution with $x<0$
Got the other part. Let $x geq y geq z geq 1$ be a solution giving $k$ that minimizes $x+y+z$ among positive solutions giving $k.$ Vieta jumping is just the new solution
$$ (x,y,z) mapsto (kyz-x,y,z) $$
The equation is
$$ x^2 - (kyz)x + y^2 + z^2 - k = 0. $$
The replacement root $x'$ gives $x + x' = kyz. $ The product is $xx' = y^2 + z^2 - k.$
We have minimized $x+y+z$ among positive triples giving $k.$ We get a new triple with $x' = kyz - x.$ As it is impossible to have $x' <0,$ we have an Alternative:
(A) $kyz - x = 0$
(B) $kyz - x geq x$
We will show that alternative (B) does not occur for, say, $k > 2.$
ASSUME (B), namely $kyz geq 2x, $ with $x geq y geq z geq 1.$
$$ x^2 + y^2 + z^2 = k + (kyz)x geq k + (2x)x= k + 2x^2 $$
$$ color{red}{ y^2 + z^2 geq k + x^2}. $$
But
$$ y^2 - (kx)yz + z^2 = k - x^2. $$
Add,
$$ 2y^2 - (kx)yz + 2z^2 geq 2k . $$
$$ y^2 - left(frac{kx}{2} right)yz + z^2 geq k . $$
So, $y geq z geq 1$ and
$$ color{red}{y^2 - left(frac{kx}{2} right)yz + z^2 > 0 }. $$
From the quadratic formula,
$$ y > frac{1}{4} left( kx + sqrt {k^2 x^2 - 16} right) z . $$
When $k geq 4,$ we see $y > x,$ a contradiction. When $k=3$ and $z geq 2,$ we still get $y > x.$ Finally, with $k=3$ and $z=1,$ the condition $y^2 + z^2 geq k + x^2$ reads $y^2 + 1 geq 3 + x^2$ or $y^2 geq 2 + x^2,$ so again $y > x.$
These contradictions tell us that alternative (A) actually holds for $k geq 3.$ That is, $kyz = x,$ so $kyz - x = 0,$ and we have a new solution $(0,y,z)$ with $y geq z geq 1$ and this $x=0$ to
$$ x^2 + y^2 + z^2 = k + kxyz. $$ As this $x=0,$ we have reached
$$ color{red}{ y^2 + z^2 = k } $$
answered Jan 31 at 21:16
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
I PROVED THE CONJECTURE
$$ x^2 + y^2 + z^2 = k + kxyz $$
Notice that we cannot have a solution with $x<0$
Got the other part. Let $x geq y geq z geq 1$ be a solution giving $k$ that minimizes $x+y+z$ among positive solutions giving $k.$ Vieta jumping is just the new solution
$$ (x,y,z) mapsto (kyz-x,y,z) $$
The equation is
$$ x^2 - (kyz)x + y^2 + z^2 - k = 0. $$
The replacement root $x'$ gives $x + x' = kyz. $ The product is $xx' = y^2 + z^2 - k.$
We have minimized $x+y+z$ among positive triples giving $k.$ We get a new triple with $x' = kyz - x.$ As it is impossible to have $x' <0,$ we have an Alternative:
(A) $kyz - x = 0$
(B) $kyz - x geq x$
We will show that alternative (B) does not occur for, say, $k > 2.$
ASSUME (B), namely $kyz geq 2x, $ with $x geq y geq z geq 1.$
$$ x^2 + y^2 + z^2 = k + (kyz)x geq k + (2x)x= k + 2x^2 $$
$$ color{red}{ y^2 + z^2 geq k + x^2}. $$
But
$$ y^2 - (kx)yz + z^2 = k - x^2. $$
Add,
$$ 2y^2 - (kx)yz + 2z^2 geq 2k . $$
$$ y^2 - left(frac{kx}{2} right)yz + z^2 geq k . $$
So, $y geq z geq 1$ and
$$ color{red}{y^2 - left(frac{kx}{2} right)yz + z^2 > 0 }. $$
From the quadratic formula,
$$ y > frac{1}{4} left( kx + sqrt {k^2 x^2 - 16} right) z . $$
When $k geq 4,$ we see $y > x,$ a contradiction. When $k=3$ and $z geq 2,$ we still get $y > x.$ Finally, with $k=3$ and $z=1,$ the condition $y^2 + z^2 geq k + x^2$ reads $y^2 + 1 geq 3 + x^2$ or $y^2 geq 2 + x^2,$ so again $y > x.$
These contradictions tell us that alternative (A) actually holds for $k geq 3.$ That is, $kyz = x,$ so $kyz - x = 0,$ and we have a new solution $(0,y,z)$ with $y geq z geq 1$ and this $x=0$ to
$$ x^2 + y^2 + z^2 = k + kxyz. $$ As this $x=0,$ we have reached
$$ color{red}{ y^2 + z^2 = k } $$
answered Jan 31 at 21:16
Will JagyWill Jagy
104k5102201
$endgroup$
I PROVED THE CONJECTURE
$$ x^2 + y^2 + z^2 = k + kxyz $$
Notice that we cannot have a solution with $x<0$
Got the other part. Let $x geq y geq z geq 1$ be a solution giving $k$ that minimizes $x+y+z$ among positive solutions giving $k.$ Vieta jumping is just the new solution
$$ (x,y,z) mapsto (kyz-x,y,z) $$
The equation is
$$ x^2 - (kyz)x + y^2 + z^2 - k = 0. $$
The replacement root $x'$ gives $x + x' = kyz. $ The product is $xx' = y^2 + z^2 - k.$
We have minimized $x+y+z$ among positive triples giving $k.$ We get a new triple with $x' = kyz - x.$ As it is impossible to have $x' <0,$ we have an Alternative:
(A) $kyz - x = 0$
(B) $kyz - x geq x$
We will show that alternative (B) does not occur for, say, $k > 2.$
ASSUME (B), namely $kyz geq 2x, $ with $x geq y geq z geq 1.$
$$ x^2 + y^2 + z^2 = k + (kyz)x geq k + (2x)x= k + 2x^2 $$
$$ color{red}{ y^2 + z^2 geq k + x^2}. $$
But
$$ y^2 - (kx)yz + z^2 = k - x^2. $$
Add,
$$ 2y^2 - (kx)yz + 2z^2 geq 2k . $$
$$ y^2 - left(frac{kx}{2} right)yz + z^2 geq k . $$
So, $y geq z geq 1$ and
$$ color{red}{y^2 - left(frac{kx}{2} right)yz + z^2 > 0 }. $$
From the quadratic formula,
$$ y > frac{1}{4} left( kx + sqrt {k^2 x^2 - 16} right) z . $$
When $k geq 4,$ we see $y > x,$ a contradiction. When $k=3$ and $z geq 2,$ we still get $y > x.$ Finally, with $k=3$ and $z=1,$ the condition $y^2 + z^2 geq k + x^2$ reads $y^2 + 1 geq 3 + x^2$ or $y^2 geq 2 + x^2,$ so again $y > x.$
These contradictions tell us that alternative (A) actually holds for $k geq 3.$ That is, $kyz = x,$ so $kyz - x = 0,$ and we have a new solution $(0,y,z)$ with $y geq z geq 1$ and this $x=0$ to
$$ x^2 + y^2 + z^2 = k + kxyz. $$ As this $x=0,$ we have reached
$$ color{red}{ y^2 + z^2 = k } $$
answered Jan 31 at 21:16
Will JagyWill Jagy
104k5102201
edited Jan 31 at 21:30
answered Jan 31 at 21:16
Will JagyWill Jagy
104k5102201
answered Jan 31 at 21:16
Will JagyWill Jagy
104k5102201
answered Jan 31 at 21:16
Will JagyWill Jagy
104k5102201
104k5102201
add a comment |
add a comment |
$begingroup$
The method is called Vieta Jumping. It is a special case of the automorphism group of an indefinite binary quadratic form.
Part Two
The first part of your question: given $k = a^2 + b^2,$ you get an answer with
$$ x=a, ; ; y =b, ; ; z = kab $$
Then
$$ frac{x^2 + y^2 + z^2}{1+xyz} = k $$
Part One
The description of Vieta Jumping in links is not quite specific enough to settle this. The conjecture seems very likely; here are the values of $k$ with $ 5000 geq a geq b geq c geq 1,$ but ignoring the many, many ways to get $k=2.$
k a b c k
5 10 2 1 5 = 5
5 1102 230 1 5 = 5
5 230 48 1 5 = 5
5 2399 48 10 5 = 5
5 48 10 1 5 = 5
5 4948 99 10 5 = 5
5 980 99 2 5 = 5
5 99 10 2 5 = 5
8 32 2 2 8 = 2^3
8 510 32 2 8 = 2^3
10 2940 297 1 10 = 2 5
10 297 30 1 10 = 2 5
10 30 3 1 10 = 2 5
10 899 30 3 10 = 2 5
13 2025 78 2 13 = 13
13 3040 78 3 13 = 13
13 78 3 2 13 = 13
17 1152 68 1 17 = 17
17 4623 68 4 17 = 17
17 68 4 1 17 = 17
18 162 3 3 18 = 2 3^2
20 160 4 2 20 = 2^2 5
25 300 4 3 25 = 5^2
26 130 5 1 26 = 2 13
26 3375 130 1 26 = 2 13
29 290 5 2 29 = 29
32 512 4 4 32 = 2^5
34 510 5 3 34 = 2 17
37 222 6 1 37 = 37
40 480 6 2 40 = 2^3 5
41 820 5 4 41 = 41
45 810 6 3 45 = 3^2 5
50 1250 5 5 50 = 2 5^2
50 350 7 1 50 = 2 5^2
52 1248 6 4 52 = 2^2 13
53 742 7 2 53 = 53
58 1218 7 3 58 = 2 29
61 1830 6 5 61 = 61
65 1820 7 4 65 = 5 13
65 520 8 1 65 = 5 13
68 1088 8 2 68 = 2^2 17
72 2592 6 6 72 = 2^3 3^2
73 1752 8 3 73 = 73
74 2590 7 5 74 = 2 37
80 2560 8 4 80 = 2^4 5
82 738 9 1 82 = 2 41
85 1530 9 2 85 = 5 17
85 3570 7 6 85 = 5 17
89 3560 8 5 89 = 89
90 2430 9 3 90 = 2 3^2 5
97 3492 9 4 97 = 97
98 4802 7 7 98 = 2 7^2
100 4800 8 6 100 = 2^2 5^2
101 1010 10 1 101 = 101
104 2080 10 2 104 = 2^3 13
106 4770 9 5 106 = 2 53
109 3270 10 3 109 = 109
116 4640 10 4 116 = 2^2 29
122 1342 11 1 122 = 2 61
125 2750 11 2 125 = 5^3
130 4290 11 3 130 = 2 5 13
145 1740 12 1 145 = 5 29
148 3552 12 2 148 = 2^2 37
170 2210 13 1 170 = 2 5 17
173 4498 13 2 173 = 173
197 2758 14 1 197 = 197
226 3390 15 1 226 = 2 113
257 4112 16 1 257 = 257
290 4930 17 1 290 = 2 5 29
k a b c k
answered Jan 31 at 2:54
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
The method is called Vieta Jumping. It is a special case of the automorphism group of an indefinite binary quadratic form.
Part Two
The first part of your question: given $k = a^2 + b^2,$ you get an answer with
$$ x=a, ; ; y =b, ; ; z = kab $$
Then
$$ frac{x^2 + y^2 + z^2}{1+xyz} = k $$
Part One
The description of Vieta Jumping in links is not quite specific enough to settle this. The conjecture seems very likely; here are the values of $k$ with $ 5000 geq a geq b geq c geq 1,$ but ignoring the many, many ways to get $k=2.$
k a b c k
5 10 2 1 5 = 5
5 1102 230 1 5 = 5
5 230 48 1 5 = 5
5 2399 48 10 5 = 5
5 48 10 1 5 = 5
5 4948 99 10 5 = 5
5 980 99 2 5 = 5
5 99 10 2 5 = 5
8 32 2 2 8 = 2^3
8 510 32 2 8 = 2^3
10 2940 297 1 10 = 2 5
10 297 30 1 10 = 2 5
10 30 3 1 10 = 2 5
10 899 30 3 10 = 2 5
13 2025 78 2 13 = 13
13 3040 78 3 13 = 13
13 78 3 2 13 = 13
17 1152 68 1 17 = 17
17 4623 68 4 17 = 17
17 68 4 1 17 = 17
18 162 3 3 18 = 2 3^2
20 160 4 2 20 = 2^2 5
25 300 4 3 25 = 5^2
26 130 5 1 26 = 2 13
26 3375 130 1 26 = 2 13
29 290 5 2 29 = 29
32 512 4 4 32 = 2^5
34 510 5 3 34 = 2 17
37 222 6 1 37 = 37
40 480 6 2 40 = 2^3 5
41 820 5 4 41 = 41
45 810 6 3 45 = 3^2 5
50 1250 5 5 50 = 2 5^2
50 350 7 1 50 = 2 5^2
52 1248 6 4 52 = 2^2 13
53 742 7 2 53 = 53
58 1218 7 3 58 = 2 29
61 1830 6 5 61 = 61
65 1820 7 4 65 = 5 13
65 520 8 1 65 = 5 13
68 1088 8 2 68 = 2^2 17
72 2592 6 6 72 = 2^3 3^2
73 1752 8 3 73 = 73
74 2590 7 5 74 = 2 37
80 2560 8 4 80 = 2^4 5
82 738 9 1 82 = 2 41
85 1530 9 2 85 = 5 17
85 3570 7 6 85 = 5 17
89 3560 8 5 89 = 89
90 2430 9 3 90 = 2 3^2 5
97 3492 9 4 97 = 97
98 4802 7 7 98 = 2 7^2
100 4800 8 6 100 = 2^2 5^2
101 1010 10 1 101 = 101
104 2080 10 2 104 = 2^3 13
106 4770 9 5 106 = 2 53
109 3270 10 3 109 = 109
116 4640 10 4 116 = 2^2 29
122 1342 11 1 122 = 2 61
125 2750 11 2 125 = 5^3
130 4290 11 3 130 = 2 5 13
145 1740 12 1 145 = 5 29
148 3552 12 2 148 = 2^2 37
170 2210 13 1 170 = 2 5 17
173 4498 13 2 173 = 173
197 2758 14 1 197 = 197
226 3390 15 1 226 = 2 113
257 4112 16 1 257 = 257
290 4930 17 1 290 = 2 5 29
k a b c k
answered Jan 31 at 2:54
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
The method is called Vieta Jumping. It is a special case of the automorphism group of an indefinite binary quadratic form.
Part Two
The first part of your question: given $k = a^2 + b^2,$ you get an answer with
$$ x=a, ; ; y =b, ; ; z = kab $$
Then
$$ frac{x^2 + y^2 + z^2}{1+xyz} = k $$
Part One
The description of Vieta Jumping in links is not quite specific enough to settle this. The conjecture seems very likely; here are the values of $k$ with $ 5000 geq a geq b geq c geq 1,$ but ignoring the many, many ways to get $k=2.$
k a b c k
5 10 2 1 5 = 5
5 1102 230 1 5 = 5
5 230 48 1 5 = 5
5 2399 48 10 5 = 5
5 48 10 1 5 = 5
5 4948 99 10 5 = 5
5 980 99 2 5 = 5
5 99 10 2 5 = 5
8 32 2 2 8 = 2^3
8 510 32 2 8 = 2^3
10 2940 297 1 10 = 2 5
10 297 30 1 10 = 2 5
10 30 3 1 10 = 2 5
10 899 30 3 10 = 2 5
13 2025 78 2 13 = 13
13 3040 78 3 13 = 13
13 78 3 2 13 = 13
17 1152 68 1 17 = 17
17 4623 68 4 17 = 17
17 68 4 1 17 = 17
18 162 3 3 18 = 2 3^2
20 160 4 2 20 = 2^2 5
25 300 4 3 25 = 5^2
26 130 5 1 26 = 2 13
26 3375 130 1 26 = 2 13
29 290 5 2 29 = 29
32 512 4 4 32 = 2^5
34 510 5 3 34 = 2 17
37 222 6 1 37 = 37
40 480 6 2 40 = 2^3 5
41 820 5 4 41 = 41
45 810 6 3 45 = 3^2 5
50 1250 5 5 50 = 2 5^2
50 350 7 1 50 = 2 5^2
52 1248 6 4 52 = 2^2 13
53 742 7 2 53 = 53
58 1218 7 3 58 = 2 29
61 1830 6 5 61 = 61
65 1820 7 4 65 = 5 13
65 520 8 1 65 = 5 13
68 1088 8 2 68 = 2^2 17
72 2592 6 6 72 = 2^3 3^2
73 1752 8 3 73 = 73
74 2590 7 5 74 = 2 37
80 2560 8 4 80 = 2^4 5
82 738 9 1 82 = 2 41
85 1530 9 2 85 = 5 17
85 3570 7 6 85 = 5 17
89 3560 8 5 89 = 89
90 2430 9 3 90 = 2 3^2 5
97 3492 9 4 97 = 97
98 4802 7 7 98 = 2 7^2
100 4800 8 6 100 = 2^2 5^2
101 1010 10 1 101 = 101
104 2080 10 2 104 = 2^3 13
106 4770 9 5 106 = 2 53
109 3270 10 3 109 = 109
116 4640 10 4 116 = 2^2 29
122 1342 11 1 122 = 2 61
125 2750 11 2 125 = 5^3
130 4290 11 3 130 = 2 5 13
145 1740 12 1 145 = 5 29
148 3552 12 2 148 = 2^2 37
170 2210 13 1 170 = 2 5 17
173 4498 13 2 173 = 173
197 2758 14 1 197 = 197
226 3390 15 1 226 = 2 113
257 4112 16 1 257 = 257
290 4930 17 1 290 = 2 5 29
k a b c k
answered Jan 31 at 2:54
Will JagyWill Jagy
104k5102201
$endgroup$
The method is called Vieta Jumping. It is a special case of the automorphism group of an indefinite binary quadratic form.
Part Two
The first part of your question: given $k = a^2 + b^2,$ you get an answer with
$$ x=a, ; ; y =b, ; ; z = kab $$
Then
$$ frac{x^2 + y^2 + z^2}{1+xyz} = k $$
Part One
The description of Vieta Jumping in links is not quite specific enough to settle this. The conjecture seems very likely; here are the values of $k$ with $ 5000 geq a geq b geq c geq 1,$ but ignoring the many, many ways to get $k=2.$
k a b c k
5 10 2 1 5 = 5
5 1102 230 1 5 = 5
5 230 48 1 5 = 5
5 2399 48 10 5 = 5
5 48 10 1 5 = 5
5 4948 99 10 5 = 5
5 980 99 2 5 = 5
5 99 10 2 5 = 5
8 32 2 2 8 = 2^3
8 510 32 2 8 = 2^3
10 2940 297 1 10 = 2 5
10 297 30 1 10 = 2 5
10 30 3 1 10 = 2 5
10 899 30 3 10 = 2 5
13 2025 78 2 13 = 13
13 3040 78 3 13 = 13
13 78 3 2 13 = 13
17 1152 68 1 17 = 17
17 4623 68 4 17 = 17
17 68 4 1 17 = 17
18 162 3 3 18 = 2 3^2
20 160 4 2 20 = 2^2 5
25 300 4 3 25 = 5^2
26 130 5 1 26 = 2 13
26 3375 130 1 26 = 2 13
29 290 5 2 29 = 29
32 512 4 4 32 = 2^5
34 510 5 3 34 = 2 17
37 222 6 1 37 = 37
40 480 6 2 40 = 2^3 5
41 820 5 4 41 = 41
45 810 6 3 45 = 3^2 5
50 1250 5 5 50 = 2 5^2
50 350 7 1 50 = 2 5^2
52 1248 6 4 52 = 2^2 13
53 742 7 2 53 = 53
58 1218 7 3 58 = 2 29
61 1830 6 5 61 = 61
65 1820 7 4 65 = 5 13
65 520 8 1 65 = 5 13
68 1088 8 2 68 = 2^2 17
72 2592 6 6 72 = 2^3 3^2
73 1752 8 3 73 = 73
74 2590 7 5 74 = 2 37
80 2560 8 4 80 = 2^4 5
82 738 9 1 82 = 2 41
85 1530 9 2 85 = 5 17
85 3570 7 6 85 = 5 17
89 3560 8 5 89 = 89
90 2430 9 3 90 = 2 3^2 5
97 3492 9 4 97 = 97
98 4802 7 7 98 = 2 7^2
100 4800 8 6 100 = 2^2 5^2
101 1010 10 1 101 = 101
104 2080 10 2 104 = 2^3 13
106 4770 9 5 106 = 2 53
109 3270 10 3 109 = 109
116 4640 10 4 116 = 2^2 29
122 1342 11 1 122 = 2 61
125 2750 11 2 125 = 5^3
130 4290 11 3 130 = 2 5 13
145 1740 12 1 145 = 5 29
148 3552 12 2 148 = 2^2 37
170 2210 13 1 170 = 2 5 17
173 4498 13 2 173 = 173
197 2758 14 1 197 = 197
226 3390 15 1 226 = 2 113
257 4112 16 1 257 = 257
290 4930 17 1 290 = 2 5 29
k a b c k
answered Jan 31 at 2:54
Will JagyWill Jagy
104k5102201
edited Jan 31 at 17:28
answered Jan 31 at 2:54
Will JagyWill Jagy
104k5102201
answered Jan 31 at 2:54
Will JagyWill Jagy
104k5102201
answered Jan 31 at 2:54
Will JagyWill Jagy
104k5102201
104k5102201
add a comment |
add a comment |
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$begingroup$
Does the usual method for the $(a^2+b^2)(ab+1)$ problem also work here?
$endgroup$
– Lord Shark the Unknown
Jan 31 at 2:48
$begingroup$
I unfortunately don't have enough experience to be able to solve that
$endgroup$
– volcanrb
Jan 31 at 2:50
1
$begingroup$
These days, this problem can be googled as well as solved: Searching for "IMO Problem 6" leads to en.wikipedia.org/wiki/Vieta_jumping
$endgroup$
– Lord Shark the Unknown
Jan 31 at 2:59
1
$begingroup$
To whoever downvoted: how can I improve the question?
$endgroup$
– volcanrb
Jan 31 at 3:11