Assume that $x\ne y$ are prime and that $y^2+y$ divides $x^2+x$. Prove that $\frac12(x-y)$ is composite.
My approach: Let's write $x^2+x$ and $y^2+y$ in a different form: $x(x+1)$, $y(y+1)$. We can now notice, that since $x$ (and also $y$) is prime and odd, then $x+1$ (as well as $y+1$) must be even. So both $x(x+1)$ and $y(y+1)$ must be even (we mutiply 2 next integers).
Now I don't know how to use a property, that $x^2+x$ is divisible by $y^2+y$.
$$\frac{x-y}{2} = \frac{(x^2+x)-(y^2+y)-(x^2-y^2)}{2} = \frac{x(x+1)-y(y+1)-(x-y)(x+y)}{2}$$
Now either
Indeed, if $y \not \mid x$ but $y(y+1) \mid x(x+1)$, then we must have either $y \mid x+1$ and $y+1 = x$, or $y \mid x+1$ and $y+1 \mid x+1$. The former case is ruled out because $x, y$ are both odd.
In the first of the two bullet points, $y$ divides everything on the numerator of the first expression. However, we can also rule out this case because $y, x$ are both prime and nonequal, so in fact we only need to look at the second case.
In the second bullet point, $y+1$ divides everything on the numerator of the last expression (indeed, it divides $x+1$ so it divides $x+1-(y+1) = x-y$).
Now, we have $y+1 \mid x-y$. If it weren't the case that $y+1 \mid \frac{x-y}{2}$, that would mean $y+1$ had exactly the same power of 2 in its prime factorisation as $x-y = (x+1)-(y+1)$ did, and hence $x+1, y+1$ have the same power of 2 in their factorisations. Therefore $\frac{x+1}{y+1}$ is odd.
But now $$\left(\frac{x-y}{2}\right)(x+y+1) = \frac{x(x+1)-y(y+1)}{2} = (y+1)\frac{x \frac{x+1}{y+1} - y}{2}$$ That final expression's numerator is therefore even, so $y+1$ is present on the RHS completely (we don't have to use up $y+1$ to absorb the 2 on the denominator). Therefore $y+1 \mid \frac{x-y}{2} (x+y+1)$.
Finally, $y+1$ is coprime with $x+y+1$ because $\text{hcf}(y+1, x+y+1) = \text{hcf}(y+1, x) = 1$.
Therefore $y+1 \mid \frac{x-y}{2}$.
How did I find this strategy? We know $y(y+1)$ divides $x(x+1)$, and we know that if $p \mid a b$ then $p \mid a$ or $p \mid b$. Therefore $y \mid x$ or $y \mid x+1$. This suggests that we need to split into two cases; but there's no good way I can see to use that $y \mid x+1$, so what can we say about the unused information that $y+1 \mid x(x+1)$? Easy: $y+1$ can't possibly divide $x$ because they aren't the same number but $x$ is prime.
