Let $\gcd(x,y)=d,x=ad,y=bd$. Obviously, $\gcd(a,b)=1$. Note that $d^2\mid d^2(a^2+b^2)\implies d^2\mid x^2+y^2$. But $x^5+y=d(a^5d^4+b)\implies d^2\mid d(a^5d^4+b)\implies d\mid b$ Similarly, $d\mid a$. Thus, $d\mid \gcd(a,b)=1\implies d=1\implies\gcd(x,y)=1$. $x^2+y^2\mid (x^5+y)(x^5-y)=(x^2)^5-y^2\implies (-y^2)^5-y^2\equiv (x^2)^5-y^2\equiv 0\pmod{x^2+y^2}$. $x^2+y^2\mid y^{10}-y^2\implies x^2+y^2\mid y^8-1$ since $\gcd(x^2+y^2,y^2)=1$. So, $x^2+y^2\mid y^8-1-y^3(y^5+x)=-(xy^3+1)$. Similarly, $x^2+y^2\mid x^8-1\implies x^2+y^2\mid x^8-1-x^3(x^5+y)=-(x^3y+1)$. $x^2+y^2\mid (x^3y+1)-(xy^3+1)=xy(x^2-y^2)$. Since $\gcd(x^2+y^2,x)=\gcd(x^2+y^2,y)=1$, $x^2+y^2\mid x^2-y^2$. If $x^2-y^2\neq 0$ then $|x^2-y^2|\geq |x^2+y^2|\implies \max(x^2,y^2)\geq x^2+y^2>\max(x^2,y^2)$ which is absurd. Thus, $x^2=y^2\implies x=\pm y$. But since $\gcd(x,y)=1$, $x,y\in\{-1,1\}$. It turn out that $(x,y)=(-1,1),(1,-1),(-1,-1),(1,1)$ all works.

My solution as I solved it in the competition: First imply $\gcd(x,y)=1$ as above. Then we have $x^2+y^2 \mid x^2+y^2+x^2(x^3y-1)\implies x^2+y^2 \mid x^3y-1$ since $\gcd(x^2,x^2+y^2)=1$ Same we get $x^2+y^2 \mid y^3x-1$. Combining these two we get $x^2+y^2 \mid 2$ and we get the solutions $(x,y)=(-1,1),(1,-1),(-1,-1),(1,1)$

Claim : $\gcd(x,y) = 1$. Proof : $\gcd(x,y) = d$ and $x = dx', y = dy', \gcd(x',y') = 1$. $d^2(x'^2 + y'^2) \mid d(d^4x' + y') \implies d \mid (d^4x' + y') \implies d \mid y'$ and $d^2(x'^2 + y'^2) \mid d(d^4y' + x') \implies d \mid (d^4y' + x') \implies d \mid x'$ so $d = 1$. $x^2 + y^2 \mid x^5 + x^3y^2 , x^2 + y^2 \mid x^5 + y \implies x^2 + y^2 \mid x^3y^2 - y \implies x^2 + y^2 \mid x(y^3x-1) \implies x^2 + y^2 \mid y^3x-1$. $x^2 + y^2 \mid y^3x + x^3y , x^2 + y^2 \mid y^3x-1 \implies x^2 + y^2 \mid x^3y - 1 \implies x^2 + y^2 \mid xy(x^2 + y^2) - 2 \implies x^2 + y^2 \mid 2$. Answers : $(x,y)=(1,1),(1,-1),(-1,1),(-1,-1)$

Good morning. I solve this problem in 2021. WLOG x>y . x^2+y^2 | x^5+y then x^2+y^2 | x^6+yx and x^2+y^2 | yx+y^6. x^2+y^2 | 2xy we have x,y>0 cause if x<0 then -x>0 we may change x and -x. 2xy>x^2+y^2 (x-y)^2<0 contraddiction.Then x=y 2x^2 |x(x^4+1) 2x | x^4+1 then sols (1,1);(1,-1)(-1,1);(-1;-1)

why post same thing 2 times? also this is a 2022 olympiad problem so u can’t have done it in 2021 lol

Lionking212 wrote: why post same thing 2 times? also this is a 2022 olympiad problem so u can’t have done it in 2021 lol Hello, some problems featured on olympiads have been tampered with in some form in the past; the problem or a similar version of it can exist well before the inclusion of it on the olympiad or the exam

oh ok i didn’t know that