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. 
 A: 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.
A: Assume $p,q$ not divisible by $5$.
$p=5k\pm1$ or $p=5k\pm2$
$$p=5k-1,5k+1,5k-2,5k+2$$
$$2p-1 = 10k-3,10k+1,10k-5,10k+3$$
But $-3$ and $+3$ are not quadratic residues modulo $5$, so $$p=5k+1, 5k-2$$
Obtain similar constraint on $q$ using fact that $2q-1$ is perfect square. Use last condition for the kill.
Thus $p$ and $q$ have to be divisible by $5$, but they are prime, so $p=q=5$
A: I'd like to offer another answer based on the unique factorization of $\mathbb{Z}[i]$.
Notice that if we let $a^2=2p-1$, $b^2=2q-1$ and $c^2=2pq-1$, we can write $$p=\left(\frac{a+1}{2}\right)^2+\left(\frac{a-1}{2}\right)^2, q=\left(\frac{b+1}{2}\right)^2+\left(\frac{b-1}{2}\right)^2$$ and $pq=\left(\frac{c+1}{2}\right)^2+\left(\frac{c-1}{2}\right)^2$. This already implies, that $a,b,c$ are odd (otherwise $p,q$ wouldn't be whole numbers) and by Fermat's theorem that $p,q$ are of the form $4k+1$ since they are sums of two squares. (Also this representation is unique).
Now let $\alpha=\frac{a+1}{2}+\frac{a-1}{2}i$, $\beta=\frac{b+1}{2}+\frac{b-1}{2}i$ and $\gamma=\frac{c+1}{2}+\frac{c-1}{2}i$. Then $\alpha$ and $\beta$ are Gaussian primes, since their norm is prime. Also, if we let $\epsilon$ be a unit ($\pm 1$ or $\pm i$) then by unique factorization of Gaussian integers, we can write $\alpha \beta=\epsilon \gamma$.
This equation leads to four cases, with two equations in each case (the real and the imaginary part). Writing out and adding/subtracting to eliminate $c$ gives the following four possibilities:
$$
(a-1)(b-1)-2 = \pm2 \\
(a+1)(b+1)-2 = \pm2
$$
By unique factorization of the integers, this leads to:
$$
a, b = \pm(1, k), \pm(k, 1), \pm(3, 3), \pm(2, 5), \pm(5, 2)
$$
Since $a=\pm1$ or $b=\pm1$ implies that one of $p$ or $q$ equals $1$ and both $a$ and $b$ need to be odd, we are left with $a,b=\pm(3,3)$ which implies $p,q=5,5$
