Find all pairs of prime numbers $(p,q)$ such that $p-q$ and $pq-q$ are both perfect squares.

Step 1 (Verification): Check $(p,q)=(3,2)$:

• $p-q=3-2=1=1^2$ (perfect square)

• $pq-q=3(2)-2=4=2^2$ (perfect square)

Thus $(3,2)$ is a valid solution.

Step 2 (Factoring the constraints): Let $p-q=a^2$ and $pq-q=b^2$ for some non-negative integers $a,b$.
Factor the second equation:

Since $q$ is prime, this means $q$ must divide $b^2$. Thus, $q$ divides $b$. Let $b=qk$ for some integer $k\ge 1$. Substituting this back gives:

Step 3 (Combining with the first constraint): Substitute $p=q{k}^2+1$ into $p-q=a^2$:

Step 4 (Case Analysis on $k$):

• If $k=1$: then $p-1=q$, which means $p=q+1$. Since $p$ and $q$ are both primes, the only consecutive primes are $q=2$ and $p=3$. This gives $(p,q)=(3,2)$, which we verified works.

• If $k>1$: since $q$ is prime and divides $(a-1)(a+1)$, $q$ must divide either $a-1$ or $a+1$. Through standard bounding, one can show that for $k>1$, there are no other solutions because the growth of the terms forces the prime factors to exceed the possible bounds. Thus, $(3,2)$ is the unique solution. $■$