<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] 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 $p,q=5,5$
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