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Source: The 1st India-Iran Friendly Competition Problem 4

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1979 posts

#1 h Jun 13, 2024, 3:45 PM • 4 Y

Y by NO_SQUARES, Blue_banana4, vexploresmathysics, sami1618

Prove that there are no integers $x, y, z$ satisfying the equation $x^2+y^2-z^2=xyz-2.$
Proposed by Navid Safaei

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#2 h Jun 13, 2024, 3:59 PM • 1 Y

Y by sami1618

Another nice problem by Mr.Safaei!
Note that $x^2+y^2=z^2+xyz-2$ , or $(x+y)^2-2=xy(z+2)+z^2-4$ and $(x-y)^2-2=xy(z-2)+z^2-4$ , so there exists integers $a,b$ such that $z+2\mid a^2-2$ and $z-2\mid b^2-2$ ( $a=x+y$ and $b=x-y$ ). Thus for every odd prime divisor of $z-2$ and $z+2$ like $p$ we have that $(\frac{2}{p})=1$ , by the following lemma we have that $p\equiv \pm 1$ (mod $8$ ). If $2\mid z$ , since $4\nmid a^2-2$ we have that $v_2(z+2)=1$ , so $z=4k$ . By the following lemma we have $8\mid 2k$ so $k=4t$ . Now $z=16t$ so $x^2+y^2\equiv 14$ (mod $16$ ) which is impossible since $x^2\equiv 1,9$ (mod $16$ ) because $x$ is odd. So $z$ is odd and by the lemma we have $(z+2,z-2)=(1,7)$ (mod $8$ ) and its impossible, thus we are done!
Lemma: For an odd prime $p$ , $(\frac{2}{p})=1\iff p\equiv \pm 1 (mod 8)$
Please correct me if I'm wrong

This post has been edited 1 time. Last edited by ayeen_izady, Jun 13, 2024, 4:00 PM
Reason: .

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#3 h Jun 13, 2024, 4:28 PM • 2 Y

Y by VicKmath7, sami1618

$z^2+xyz-x^2-y^2-2=0$ $\Delta=(xy)^2+4x^2+4y^2+8$
$t^2=x^2y^2+4x^2+4y^2+8\iff t^2+8=(x^2+4)(y^2+4)$ If both $x,y$ are even, then $16|t^2+8$ which is impossible. WLOG let $x$ be odd. Take a prime $p|x^2+4$ . We have $(\frac{-1}{p})=(\frac{-4}{p})=1$ hence $p\equiv 1(mod \ 4)$ . Also $p|t^2+8$ gives $(\frac{2}{p})=(\frac{-2}{p})=(\frac{-8}{p})=1$ so $p\equiv 1,7(mod \ 8)$ . These two give that $p\equiv 1(mod \ 8)$ but since $x^2+4 \not \equiv 1(mod \ 8),$ there exists a prime $\not \equiv 1(mod \ 8)$ . Thus, we get contradiction as desired. $\blacksquare$

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L567

#4 h Jun 14, 2024, 7:34 AM • 1 Y

Y by sami1618

Suppose there exists such a triple. It's easy to check that none of them can be zero. Note we can flip signs of any two and still keep equation valid, so wlog say $x, y > 0$ . Note that if $z$ works, then $\frac{-2-x^2-y^2}{z}$ also works (and is an integer, by vieta), so we can also assume that $z > 0$ . Now consider a triple with all $x,y,z$ positive with $x+y$ minimal. But now, note that (wlog $x \geqslant y$ ) if $x$ works, by vieta so does $\frac{y^2 - z^2 + 2}{x}$ . Since $x$ was minimal, we have that $\frac{y^2 - z^2 + 2}{x} \geqslant x$ or that $x^2 \geqslant y^2 \geqslant x^2 + z^2 - 2$ so we must have $z = 1$ , but again in this case it's easy to see there are no solutions.

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#5 h Jun 14, 2024, 10:50 AM • 1 Y

Y by sami1618

L567 wrote:

if $x$ works, by vieta so does $\frac{y^2 - z^2 + 2}{x}$ .

Couldn't it happen that this second solution is negative? (And then you can't continue to argue by minimality...)

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#6 h Jun 19, 2024, 8:06 PM

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We can view this as a quadratic in $z$ . The discriminant of this equation must be a perfect square, say $s^2$ .
$\begin{align*} \Delta=s^2 & =x^2y^2+4x^2+4y^2+8 \\ s^2 & =(x^2+4)(y^2+4)-8 \\ s^2+8 & =(x^2+4)(y^2+4) \end{align*}$ It is standard to show that if some odd prime $p$ divides $s^2+8$ then $p\equiv 1,3 \pmod{8}$ and if some odd prime $p$ divides $x^2+4$ then $p\equiv 1,5 \pmod {8}$ . Thus for all odd prime factors $p$ of both sides $p\equiv 1 \pmod{8}$ . Now notice that if both $x$ and $y$ are even then $16|(x^2+4)(y^2+4)$ but it is not hard to show that $16 \nmid (x^2+4)(y^2+4)$ . So assume that $x$ is odd. Then $x^2+4\equiv 3,5 \pmod{8}$ , a contradiction.

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#7 h Sep 27, 2024, 7:43 AM

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Let x•y = even no. (Like 2,4,6,8,...)
Now after you solve the equation considering xy=even then it's never possible to get such x,y,z integers. Let now consider x•y=odd no. Which again proves it's impossible

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