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$x,y,z \in \mathbb{Z^+}$ such that $x^2+y^2=z^2(1+xy)$. Prove $z=\min \{x;y;z\}$

$$x^2+y^2=z^2(1+xy) \iff xy = \frac{x^2+y^2} {z^2} - 1$$. Assum $z>y \implies xy < x^2/z^2$, we have $xy \in Z \implies x> z$ then...stuck

Michael Hardy
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Xeing
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  • Isn't the definition of $z$ circular? – Pedro Feb 14 '13 at 15:45
  • Sorry, I misread the problem. I thought $z$ was supposed to be unknown! – Pedro Feb 14 '13 at 16:07
  • possible duplicate of $\frac{(a^2+b^2)}{(1+ab)}$ must be a perfect square if it is an integer – Gerry Myerson Feb 15 '13 at 04:53
  • Which in turn was closed as a duplicate of http://math.stackexchange.com/questions/28438/alternative-proof-that-a2b2-ab1-is-a-square-when-its-an-integer – Gerry Myerson Feb 15 '13 at 04:54
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    This is a different problem. We are not asked to show that $z$ is an integer. We are given that $z$ is an integer and shall prove $z\le x$, $z\le y$. – Hagen von Eitzen Feb 15 '13 at 07:38
  • @Hagen, yes, but the way you solve the other problem(s) is to figure out what $x$ and $y$ have to be, and then it's immediate that $z^2$ is the square of the smaller of $x$ and $y$. – Gerry Myerson Feb 15 '13 at 12:09
  • Is there any way to find $a, b \in \mathbb{N}$ such that $1+ab \mid a^2 + b^2$ – Xeing Feb 15 '13 at 12:09
  • @Gerry, I would (and did, below) call this a variant, not a duplicate. The $ab+1$ problem, restated in the terms of this question, essentially is to show that $(0,z)$ is the smallest integer solution. This question asks for the smallest positive integer solution, which is $(z,z^3)$. – zyx Feb 16 '13 at 10:28
  • @D3r0X4, the solution $(a,b) = (z,z^3)$ was written in the answer and comments. There are many others. – zyx Feb 16 '13 at 10:30
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    This should not be closed as a duplicate. It contains the non-Diophantine problem of seeing that a proof by pure inequalities cannot work. The Diophantine part is similar to, but somewhat more complicated than, and requires a bit more understanding of the structure, than the older questions on ab+1. – zyx Feb 16 '13 at 11:10

1 Answers1

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This is a variant of the $ab+1$ problem (an old and famous killer problem from the IMO), and is a corollary of its solution. For fixed $z$, the solutions $(x,y)$ organize into a chain where the size of the solution can be reduced by moving down the chain until the minimal solution with $xy=0$ is reached. This smallest solution is $(0,z)$ or $(z,0)$. The smallest positive solution is adjacent to this one in the chain, and adjacent solution have one variable the same, so the smallest positive solution is $(z,t)$ or $(t,z)$ for some value of $t$ (larger than $z$, in fact $t = z^3$). This has $\min(x,y)=z$ and the other positive solutions have bigger values of $\min(x,y)$. For the same reasons, in a positive solution, $\max(x,y) \geq z^3$. The inequalities are a complicated way of saying that $(x,y)=(z,z^3)$ is the smallest positive integer solution to the equation.

There was a careful analysis of the $ab+1$ problem here:

Seemingly invalid step in the proof of $\frac{a^2+b^2}{ab+1}$ is a perfect square?

Edit:

Actually the reduction process, applied to any non integer solution with fixed $z$, should eventually reach a pair $(x,y)$ with $0 < x < z \leq y$, so this problem can work only as a Diophantine question and is not derivable from inequalities. The minimal solution overall will have $xy < 0$ and both variables less than $z$ in absolute size.

Community
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zyx
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