maybe a start: note that the terms are in arithmetic progression

what competition is that?

i think it is the international olympiad of metropolises which was held this year at Moscow

Severus wrote: i think it is the international olympiad of metropolises which was held this year at Moscow thanks who is allowed to participate in it?

Medjl wrote: Severus wrote: i think it is the international olympiad of metropolises which was held this year at Moscow thanks who is allowed to participate in it? You could check this out. Might help http://megapolis.educom.ru/en/about-faq

We'll not consider the case when $x=y$. For each $i\in \{ 0,1,2\}$, let $d_i=\gcd (x+i,y+i)$. Not hard to see that $d_0,d_1,d_2$ are positive divisors of $|x-y|$ and that $\gcd (d_0,d_1)=\gcd (d_1,d_2)=1,\gcd (d_0,d_2)\mid 2$. We've $\frac{(x+2)(y+2)}{d_2}+\frac{xy}{d_0}=\frac{2(x+1)(y+1)}{d_1}\Rightarrow xy(d_0d_1+d_1d_2-2d_0d_2)+x(2d_0d_1-2d_0d_2)+y(2d_0d_1-2d_0d_2)+(4d_0d_1-2d_0d_2)=0$. So $\Big( x(d_0d_1+d_1d_2-2d_0d_2)+(2d_0d_1-2d_0d_2)\Big) \Big( y(d_0d_1+d_1d_2-2d_0d_2)+(2d_0d_1-2d_0d_2)\Big) =2d_0d_1d_2(d_0+d_2-2d_1)$. Not hard to show that all of $d_0,d_1,d_2$ divide both $x(d_0d_1+d_1d_2-2d_0d_2)+(2d_0d_1-2d_0d_2)$ and $y(d_0d_1+d_1d_2-2d_0d_2)+(2d_0d_1-2d_0d_2)$. Note that the least common multiple of $d_0,d_1,d_2$ is equal to $d_0d_1d_2$ or $\frac{d_0d_1d_2}{2}$, while the latter case is when $d_0$ and $d_2$ are even. Both cases give $d_0d_1d_2\mid 4(d_0+d_2-2d_1)$. Since $4(d_0+d_2-2d_1)\leq 4(d_0+d_2-2)\leq 4d_0d_2$ and $4(d_0+d_2-2d_1)\geq 4(2-2d_1)$. If $4(d_0+d_2-2d_1)=0\Rightarrow d_0+d_2=2d_1$, we get, WLOG it's $x$, $x(d_0d_1+d_1d_2-2d_0d_2)+(2d_0d_1-2d_0d_2)=0$. So, if $d_2\neq d_1$, which otherwise leads to $d_0=d_1=d_2=1\Rightarrow 4-2=0$, contradiction, we must have $x=\frac{2d_0d_2-2d_0d_1}{d_0d_1+d_1d_2-2d_0d_2}$. Plug in $d_2=2d_1-d_0$ gives us $x=\frac{d_0}{d_1-d_0}\Rightarrow d_1-d_0\mid \gcd (d_0,d_1)=1\Rightarrow d_1-d_0=1\Rightarrow x=d_0$. Hence $x=d_0=\gcd (x,y) \mid y$, done. If $4(d_0+d_2-2d_1)\neq 0$. Either $4(d_0+d_2-2d_1) \geq d_0d_1d_2$ or $4(d_0+d_2-2d_1)\leq -d_0d_1d_2$. When $4(d_0+d_2-2d_1) \geq d_0d_1d_2$, we get $4d_0d_2\geq d_0d_1d_2\Rightarrow d_1\leq 4$. When $4(d_0+d_2-2d_1)\leq -d_0d_1d_2$, we get $-d_0d_1d_2\geq 8-8d_1\Rightarrow 7\geq d_0d_2$. The rest is, I believe, only long cases bashing which I'll leave it to the reader.

$x\ge y$ $\frac{(x+2)(y+2)}{(x+2,y+2)}-2\frac{(x+1)(y+1)}{(x+1,y+1)}+\frac{xy}{(x,y)}=0$ $\frac{y+2}{(x+2,y+2)}-\frac{y}{(x,y)}\equiv 0\pmod{x+1}$ $\frac{y+2}{(x+2,y+2)}=\frac{y}{(x,y)}$ $\frac{(x+2)y}{(x,y)}-2\frac{(x+1)(y+1)}{(x+1,y+1)}+\frac{xy}{(x,y)}=2\frac{(x+1)y}{(x,y)}-2\frac{(x+1)(y+1)}{(x+1,y+1)}=0$ $\frac{y}{(x,y)}=\frac{y+1}{(x+1,y+1)}$ $x=dp,y=dq,q=\frac{dq+1}{(x+1,dq+1)},q(x+1,dq+1)=dq+1$ $q|1,q=1,d=y,y|x$

$[x+2,y+2]$ + $[x,y]$ = $2[x+1,y+1]$ WLOG, assume $x > y$ let $[x+2,y+2]$ = $k.(x+2)$ and $[x,y]$ = $lx$. By the assumption, we know $0<k<x+2$ and $0<l<x$ Then we have $x+1 | k.(x+2)+ lx$ $=>$ $x+1 | k - l$. $=>$ $|k-l| \ge x+1$ or $k=l$. But due to the bounds of $k$ and $l$ above, we have the latter: $k=l$. Note that $k =\frac{(y+2)}{(x+2,y+2)}$, So $k|y+2$. Similarly, $l=k|y$ $=>$ $k | 2$. This narrows down the problem to $k=2$ or $k=1$. If $k=2$, then $2|y$ $=>$ $y+1$ is odd. Subsituting this to the equality gives us $[x+1,y+1]=2(x+1)$, which is not possible due to the $y+1$ being odd. So $k=1$. But this results in $y+2 | x+2$ and $y|x$, and we are done! If we have the minor case $x=y$, we are still done.