Solution Verification: 1986 IMO Problem 1

Question

I'm very new to IMO problems but I'm extremely interested in learning how to solve these problems. The problems that I'm working on have solutions available on the internet, and though I've consulted some outside sources, I'm very interested in being able to understand every step and justify everything rigorously in my own words. Therefore, I would very much appreciate if someone could critique my attempt to solve this problem.

The Problem

Let $d$ be any positive integer not equal to $2$, $5$, or $13$. Show that one can find distinct $a,b$ in the set $\{2,5,13,d\}$ such that $ab - 1$ is not a perfect square.

My Attempt at a Solution

For the sake of contradiction, let us suppose that for all distinct $a,b$ in this set, $ab - 1$ is a perfect square. In particular, \begin{align*} 2d - 1 = x^2, \qquad 5d - 1 = y^2, \qquad 13d - 1 = z^2 \end{align*} for $x,y,z \in \mathbb{Z}$. As $2d - 1$ is odd, $x^2$ must be odd and hence $x$ is odd; if $x$ were even, its square would be even. One can then write $x = 2k - 1$ for some $k \in \mathbb{Z}$. Substituting then yields \begin{align*} & 2d - 1 = (2k-1)^2 \\ & 2d - 1 = 4k^2 - 4k + 1 \\ & d = \frac{4k^2 - 4k + 2}{2} = 2k^2 - 2k + 1 = 2(k^2 - k) + 1, \end{align*} so $d$ is odd. Therefore, $5d$ and $13d$ are products of odd numbers and hence odd, so $5d - 1$ and $13d - 1$ are even, hence $y^2$ and $z^2$ are even, hence $y$ and $z$ are even. Let $y = 2a$ and $z = 2b$ for some $a,b \in \mathbb{Z}$. Substituting then yields \begin{align*} & 5d - 1 = (2a)^2 = 4a^2 \\ & 13d - 1 = (2b)^2 = 4b^2. \end{align*} Subtracting the first equation from the second, one obtains \begin{align*} & (13d - 1) - (5d - 1) = 4b^2 - 4a^2 \\ & 13d - 5d = 4(b^2 - a^2) \\ & 8d = 4(b+a)(b-a) \\ & 2d = (b+a)(b-a). \end{align*} As $d$ is odd, $2 \nmid d$, so $2 \mid 2d$ but $4 \nmid 2d$, that is, $2d \not \equiv 0 \pmod 4$. However, $b + a$ and $b - a$ have the same parity, i.e., $b + a$ is even if and only if $b - a$ is even. If $b + a$ and $b-a$ are both odd, their product is odd and hence not divisible by $2$, which is a contradiction. Therefore, they must both be even, meaning that they are each divisible by $2$ and hence their product is divisible by $4$, so $(b+a)(b-a) \equiv 0 \pmod 4$. Hence, $2d \not \equiv (b+a)(b-a) \pmod 4$, so $2d \neq (b+a)(b-a)$, a contradiction. Consequently, it is not the case that for al $a,b$ in the aforementioned set, $ab - 1$ is a perfect square, as required.

I did not include in my write-up a proof that $b + a$ and $b - a$ have the same parity, as I wasn't sure if that type of statement could be taken for granted. Regardless, here is the proof I alluded to above. I'd also appreciate if someone could critique this argument if there is something wrong with it.

It suffices to prove that $b + a$ is even if and only if $b-a$ is even. Suppose $b + a$ is even, so $b + a = 2k$ for some integer $k$. We then have $$ b - a = (b+a) - 2a = 2k - 2a =2(k-a), $$ so $b$ is even. If instead $b - a$ were even, then $b - a = 2n$ for some $n$, and one has $$ b + a = (b-a) + 2a = 2n + 2a = 2(n+a), $$ so $b+a$ is even.