Silk Road 2021 P4

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#1 h Jun 28, 2021, 6:23 PM

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Integers $x,y,z,t$ satisfy $x^2+y^2=z^2+t^2$ and $xy=2zt$ prove that $xyzt=0$

Proposed by $M. Abduvaliev$

This post has been edited 4 times. Last edited by Hopeooooo, Jan 8, 2023, 10:50 PM
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#2 h Jun 28, 2021, 6:45 PM • 2 Y

Y by InvertedDiabloNemesisXD, Danie1

$x^2+y^2+xy=(z+t)^2\Longrightarrow xy=0$

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30627 posts

#3 h Jun 28, 2021, 8:15 PM

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@Hopeoooo You might add the proposer's name

M. Abduvaliev

This post has been edited 1 time. Last edited by parmenides51, Jun 28, 2021, 8:15 PM

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Jjesus

#4 h Jun 29, 2021, 10:20 PM • 1 Y

Y by oolite

ratatuy wrote:

$x^2+y^2+xy=(z+t)^2\Longrightarrow xy=0$

this because?

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kvanta

#5 h Jun 30, 2021, 2:52 AM

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+1 to the above: $x=3, y=5$ gives a square so this needs way more justification

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kvanta

#6 h Jun 30, 2021, 2:56 AM

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http://matol.kz/comments/4559/show
Here's a link to the official solution. Should still make sense even if you don't understand Russian or Kazakh

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kvanta

#7 h Jul 5, 2021, 6:44 AM • 2 Y

Y by oolite, Syvl

Here's the solution in English anyway (solution by A. Golovanov):

Assume that $xyzt\ne 0$ .

If three of the numbers $x$ , $y$ , $z$ , $t$ have a common divisor $d$ , the
remaining number is also a multiple of $d$ . The division of all the four numbers
by $d$ preserves the validity of our equations. Since $x$ , $y$ , $z$ , $t$ are not
zero, we can divide them by their greatest common divisor and henceforth assume
that they are coprime.

The numbers $x$ and $y$ can not be both odd (since $xy$ is even) or both even
(since in this case $zt$ is also even and at least three of the numbers $x$ ,
$y$ , $z$ , $t$ are even). Thus $x$ and $y$ and therefore $z$ and $t$ have
different parity.

Denote $s=x^2+y^2$ , $n=zt$ . Without loss of generality, $n > 0$ .

The condition on $x$ , $y$ , $z$ , $t$ means that $s-4n$ , $s-2n$ , $s+2n$ , $s+4n$
are perfect squares. We will prove that this is not possible for any odd $s$ and
positive $n$ . First, we find a statement equivalent to these four numbers being
squares. Their product $(s^2-4n^2)(s^2-16n^2)=(s^2-10n^2)^2-36n^4$ is a square
of some odd $T$ . The numbers $T$ , $6n^2$ , $s^2-10n^2$ form a primitive
Pythagorean triple, hence there exist coprime $u$ , $v$ of different parity such
that $2uv=6n^2$ and $u^2+v^2=s^2-10n^2$ . Since $u$ and $v$ are coprime, it
follows from $uv=3n^2$ that one of these numbers is a square and another is a
triple square. Without loss of generality, assume that $u=3k^2$ and $v=\ell^2$ ;
$k$ and $\ell$ are coprime. Multiplying $u^2+v^2=s^2-10n^2$ by 3 we get
$3s^2-30n^2=3s^2-10uv=3u^2+3v^2$ , that is, $3s^2=3u^2+10uv+3v^2=(3u+v)(3v+u)$ ,
where $3u+v$ and $3v+u$ are coprime (they have different parity, and their
common divisor must also divide $8u=3(3u+v)-(3v+u)$ and, similarly, $8v$ ). It
follows that one of the numbers $3u+v$ and $3v+u$ is a square and another is a
triple square. Since 3 does not divide $3u+v$ , the square is $3u+v=9k^2+\ell^2$ ,
and the triple square is $u+3v=3(k^2+\ell^2)$ . We have proved that the original
equations imply the existence of coprime $k$ and $\ell$ such that $9k^2+\ell^2$
and $k^2+\ell^2$ are perfect squares.

It remains to prove that there are no such $k$ , $\ell$ . If $k^2+\ell^2$ and
$9k^2+\ell^2$ are perfect squares, the pairs $(k, \ell)$ and $(3k, \ell)$ are
pairs of legs in primitive Pythagorean triples. Unfortunately, we do not know
which of the numbers $k$ and $\ell$ is even, so two cases are possible. But
let's prove the following lemma first.

Lemma: Let $a, b, c, d$ be pairwise coprime positive integers such that $c^2 (a^2 + d^2) = b^2 (9a^2 + d^2)$ . Then, $a^2 + d^2$ and $9a^2 + d^2$ are perfect
squares.

Proof: Since $GCD(b, c) = 1$ , then there exists such positive integer $f$ that
$a^2 + d^2 = b^2 f, \quad 9a^2 + d^2 = c^2 f.$ $f = GCD(a^2 + d^2, 9a^2 + d^2) \implies f \ | \ 8 a^2, \quad f \ | \ 8 d^2.$ $GCD(a, d) = 1 \implies f \ | \ 8.$ If $a, d$ are of different parity, then
$a^2 + d^2$ is odd, and therefore $f = 1$ . Let both $a$ and $d$ be odd (they
can't be simultaneously even). Then
$a^2 + d^2 \equiv 2 \pmod{8} \implies f \ | \ 2.$ Let $f = 2$ .
$9a^2 + d^2 = c^2 f = 2 c^2 \implies d^2 \equiv 2 c^2 \pmod{3} \implies 3 \ | \ c, \quad 3 \ | \ d$ --- a contradiction, since $GCD(c, d) = 1$ . Thus, $f = 1$ , $a^2 + d^2 = b^2$ , $9a^2 + d^2 = c^2$ . The lemma is proven.

Case 1. $k$ is even. Then $k=2u_1v_1$ , $3k=2u_2v_2$ ,
$\ell=u_1^2-v_1^2=u_2^2-v_2^2$ , where $(u_1, v_1)$ and $(u_2, v_2)$ are two
pairs of coprime numbers with different parity. Since $u_2v_2=3u_1v_1$ , then
there exist pairwise coprime positive integers $a$ , $b$ , $c$ , $d$ such that
$u_2=3ab$ , $v_2=cd$ , $u_1=ac$ , $v_1=bd$ (the case when $u_2$ is not divisible by
3 can be considered similarly). Substituting this in the expression for $\ell$
we obtain $a^2c^2+c^2d^2=9a^2b^2+b^2d^2$ , or $c^2(a^2+d^2)=b^2(9a^2+d^2)$ .
According to the lemma, $a^2+d^2$ and $9a^2+d^2$ are squares, and, obviously,
$a<k$ , $d<\ell$ .

Case 2. $\ell$ is even. Then $k=u_1^2-v_1^2$ , $3k=u_2^2-v_2^2$ ,
$\ell=2u_1v_1=2u_2v_2$ . Since $(u_2-v_2)(u_2+v_2)=3(u_1-v_1)(u_1+v_1)$ , then we
have $u_1-v_1=ab$ , $u_1+v_1=cd$ , $u_2-v_2=ac$ , $u_2+v_2=3bd$ for some pairwise
coprime positive integers $a$ , $b$ , $c$ , $d$ (the case when $u_2 + v_2$ is not
divisible by 3 can be considered similarly). Expressing $u_1$ , $v_1$ , $u_2$ ,
$v_2$ in terms of $a$ , $b$ , $c$ , $d$ and putting the result in the formula for
$\ell$ , we get $(ab+cd)(cd-ab)=(3bd+ac)(3bd-ac)$ , that is,
$c^2(a^2+d^2)=b^2(a^2+9d^2)$ . Hence, by the lemma, we obtain that $a^2+d^2$ and
$a^2+9d^2$ are perfect squares, and $a<\ell$ , $d<k$ .

In both cases, for each pair $(k, \ell)$ such that $k^2+\ell^2$ and
$9k^2+\ell^2$ are perfect squares we constructed a pair of smaller numbers with
the same property. It follows that such numbers do not exist. Thus, the original
system does not admit a solution in non-zero integers.

This post has been edited 1 time. Last edited by kvanta, Jul 5, 2021, 6:49 AM
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