<s>[INST] Finding all primes $(p,q)$ for perfect squares. Find all prime pairs $(p,q)$ such that $2p-1, 2q-1, 2pq-1$ are all perfect squares.
Source: St.Petersburg Olympiad 2011
I have only found the pair $(5,5)$ so I am thinking that maybe a modulo $5$ approach could work. 
 [/INST] Although Ivan Loh's answer is correct, I believe I may have found a much simpler and elementary proof that $(5,5)$ is the only pair. 
Assume without loss of generality that $q\geq p$ and write $2pq-1 = x^2$, 
$2q-1 = y^2$, with $x$ and $y$ positive integers which are necessarily both odd.
We have the following inequalities.
$$x = \sqrt{2pq-1} < \sqrt{2pq} \leq q\sqrt{2}$$
$$y = \sqrt{2q-1} < \sqrt{2q}$$
For the first inequality we have used the assumption that $p\leq q$. Now we have that
$$x^2 - y^2 = (2pq-1) - (2q-1) = 2q(p-1)$$
So $2q(p-1) | (x+y)(x-y)$. Since $q$ is prime, we must now have either
$q|(x-y)$ or $q|(x+y)$. Additionally, $x$ and $y$ are both odd, so
$x-y$ and $x+y$ are both even. Combining this with the preceding statement we see that either $2q|(x-y)$ or $2q|(x+y)$.
The former cannot occur since $x-y < x < q\sqrt{2} < 2q$, so if 
$2q|(x-y)$ then $x = y$ which would imply that $p = 1$.
So $2q|(x+y)$. Now $x + y > 0$ so we must have
$$2q \leq x + y < q\sqrt{2} + \sqrt{2q}$$
Rearranging this we get
$$(2-\sqrt{2})q < \sqrt{2q}$$
so
$$(2-\sqrt{2})^2q < 2$$
and so $q < \frac{2}{(2-\sqrt{2})^2} < 5.8$. Since both $p$ and $q$ are $1$ mod $4$, this leaves $p = q = 5$ as the only possible pair.
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