<s>[INST] $\frac{1}{2}(x-y)$ is prime where $x,y$ are distinct odd primes and $y^2 + y$ divides $x^2+x$ I need help with the following exercise from the Polish XI Olimpiada Matematyczna Gimnazjalistów (question 6 from this document. The submission deadline of 12 October 2015 has passed.

Assume that $x≠y$ are both prime, odd and that $y^2+y$ divides $x^2+x$. Prove that $\frac{1}{2}(x−y)$ is composite.

I think I'm really close to find a solution. My approach takes advantage of the Euclidean algorithm:
$\gcd(x^2+x, y^2+y) = \gcd(x^2-y^2+x-y, y^2+y)=\gcd((x-y)(x+y)+(x-y), y^2+y)=\gcd((x-y)(x+y+1),y^2+y)=y^2+y$
Now it's true, that:
$\frac{(x-y)(x+y+1)}{(y^2+y)}=a$, where $a$ is natural greater than $1$
$(x-y)(x+y+1)=a(y^2+y)$
$\frac{1}{2}(x-y)=\frac{a(y^2+y)}{2(x+y+1)}$
I think now I should prove, that $\frac{a(y^2+y)}{2(x+y+1)}$ is composite, but I don't know how to.
 [/INST] You are given 
$$y(y+1)\mid x(x+1) \tag{1}$$
so the fact that $y\ge3$ and $y\ne x$ implies that $y<x$. Since $x,y$ are both odd primes $y\le x-2$ and $y\ge3$.
The since $x$ is prime and $y<y+1<x$ we must have  
$$\begin{array}{ll}
\gcd(y,x)=1\text{ and }\gcd(y+1,x)=1 \\
\implies \color{blue}{y(y+1)\mid x+1} &(\text{from (1)}) \\
\implies x+1=ky(y+1),\quad k\ge1,k\in\mathbb{Z}
\end{array}$$
So $\frac{1}{2}(x-y)=\frac{1}{2}(ky-1)(y+1)=(ky-1)(\frac{y+1}{2})$. Since $k\ge1,y\ge3$ have that $ky-1\ge2$ is an integer, and $y\ge3$ and odd means that $\frac{y+1}{2}\ge2$ is an integer, so $\frac{1}{2}(x-y)$ has two integer factors of at least two, so cannot be a prime.

A solution to (1) is $x=11,y=3$ from which we see that $y(y+1)=12,x+1=12 \implies k=1$, so then $ky-1=2$ and $\frac{y+1}{2}=2$.  
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