Find all two-digit prime number pairs $p$ and $q$, for which $pq+1$ is a perfect square.

Question

Find all two-digit prime number pairs p and q, for which $p\cdot q+1$ is a perfect square.

MY IDEAS

Because p and q are two digit prime numbers, clearly, they are odd numbers, so let $p=2a+1$ and $q=2b+1$ where $a, b$ are positive integers.

$$(2a+1)(2b+1)+1=(2k+1)^{2}+1$$ for some positive integer $k$.

Then we can apply the formula that says $(x+y)^2=x^2+y^2+2xy$.

But what should I do next? Hope one of you can help me! Thank you in advance!

Answer 1

We can assign $pq+1$ to equal to another variable, say $k^2$.

in other words, let $$pq+1=k^2$$ where $k$ is a positive integer.

Rearranging this equation yields $$pq=k^2 - 1$$ $$\implies pq = (k-1)(k+1)$$

We know that $p$, $q$ are prime, so the LHS should only have factors $1, p, q, pq$. Hence the RHS, $(k-1)(k+1)$ should also only have the above $4$ factors. In particular, since $p, q$ are two digit primes, so it can't be that $k-1 = 1$ and $k+1 = pq$. Hence $$p=k-1$$ $$q=k+1$$ or the other way around.

So to sum up, the answer to your question should be that $q=p+2$, or in other words, two $2$ digit primes that differ by $2$. Hopefully this helped.

Answer 2

Any pair of twin primes will work, because $(n+1)(n-1)=n^2-1$. Those are the only possible solutions because the problem is equivalent to saying $n^2-1=(n+1)(n-1)$ is a product of two primes, so those two primes must be $n-1$ and $n+1$.