India — Team Selection Test, 2007

$$y^2 = (x-p)(x)(x+p).$$
We split the analysis into the following cases.

Case $p \nmid y$. In this case, we have $(x-p,x) = (x,x+p) = 1$. If $x$ is even, then $(x-p,x+p) = 1$, so it follows that $x-p,x$ and $x+p$ are all squares. This is not possible since $x+p \equiv 3 \pmod{8}$. Thus, $x$ has to be odd. In this case we have $(x-p,x+p) = 2$. So we get
$$\begin{aligned} x &= r^2, \\ (x-p) &= 2s^2, \\ (x+p) &= 2t^2. \end{aligned}$$
The last two equations imply $(t^2 - s^2) = p$ solving which we get $t = (p+1)/2$ and $s = (p-1)/2$. Now,
$$\begin{aligned} r^2 &= x = (x-p)+p \\ &= (p-1)^2/2 + p \\ &= (p^2+1)/2 \\ &= ((8k+3)^2 + 1)/2 \quad (\text{here } p = 8k+3) \\ &= (64k^2 + 48k + 10)/2 \equiv 5 \pmod{8} \end{aligned}$$
which is not possible. Thus there are no solutions in case $p \nmid y$.

Case $p \mid y$. We first note that if $y=0$ then $x=0, p$ or $-p$, and this gives us three solutions. Now, assume that $y \neq 0$. Since $p \mid y$ it follows that $(x-p,x,x+p) = p$. Thus, $p^2 \mid y$. Canceling $p^3$ from both the sides, we have
$$pb^2 = (a-1)(a)(a+1),$$
where $a = x/p$ and $b = y/p^2$. Note that $a-1, a$ and $a+1$ are all positive since $y \neq 0$. Now, $p$ divides exactly one of $(a-1), a$ and $(a+1)$. If $p$ divides $(a-1)$ then, $a$ and $(a+1)$ have to be squares, which is not possible since $a \neq 0$. So, $p \nmid (a-1)$. Similarly, $p \nmid (a+1)$. Thus, it should be the case that $p \mid a$.
If $a \neq 0$ is even, then $(a-1), a/p$ and $(a+1)$ are relatively prime to each other, so they all have to be squares. But it is not possible that $(a-1)$ and $(a+1)$ are both squares. Thus, $a$ should be odd.
Now, $(a-1, a+1) = 2$, so $(a-1) = 2r^2$ and $(a+1) = 2s^2$. This gives $s^2 - r^2 = 1$ which implies $s=1$ and $r=0$. This gives $a=1$ which is not divisible by $p$. Thus there are no solutions in this case.