Prime number and square problem How many pairs of natural numbers, not bigger than 100, are such that difference between that pair is a prime number, and their product is a square of a natural number.
My attempt: I tried writing relationship such as $x-y=p$ and $xy=n^2$ but I can't seem to find any pattern to enumerate it.  
 A: OK, if you have $x-y=p$, then either both $x$ and $y$ are divisible by $p$, or both are not.
If they are not, then $x$ and $y$ are coprime, and their product being a square, this means they must both be squares themselves: $x=i^2,\;y=j^2$, in which case their difference factors nicely to $(i-j)(i+j)$ and can only be a prime if $i=j+1$. As Ethan noted, there are not awfully many values to check.
If they are, then $x\over p$ and $y\over p$ are two integers that differ by 1 and hence are also coprime. Notice that their product is $n^2\over p^2$, that is, also a square, so they must be squares themselves, but two squares rarely differ by 1.
So it goes.
A: If $x - y$ is prime, then there can be at most one number that divides both $x$ and $y$, and it has to be that prime. Suppose the prime $p$ divides both $x, y$. Thus we can write $x = np$, and $y = (n-1)p$. Then we would have that $n(n-1)p^2$ is a square, hence so is $n(n-1)$. But $n, n-1$ are coprime, so this would mean that both $n, n-1$ are squares. This is impossible, so $p$ does not divide $x$ or $y$. It follows that $x, y$ are coprime.
Because $xy$ is also a square, it follows that $x, y$ must be squares individually -- so write $x = a^2, y = b^2$. Now a square is a sum of odd numbers,
$$
a^2 = \sum_{k=1}^a 2k-1,
$$
so
$$
a^2 - b^2 = \sum_{k=b+1}^a 2k-1.
$$
But we run into a restriction: if we have a sequence of odd numbers of more than one term, say $5, 7, 9$ or $25, 27$, then they are a multiple of their average. The first sum equals $3 \times 7$ and the second one $2 \times 26$. Thus we must have that the sum consists of at most one term, and we have $a = b+1$, and the sum is prime if and only if $2a-1$ is prime.
In the end your numbers are the $x, y$ such that $x = \left(\frac{p+1}{2}\right)^2$ and $y = \left(\frac{p-1}{2}\right)^2$, where $p$ is an odd prime.
A: You may proceed by reducing it into $\pmod 4$. i.e. $n^2 \equiv 0,1 \pmod 4$ and $p \equiv 1,3 \pmod 4$. So, the reduced problem is $$xy\equiv 0,1 \pmod 4$$
$$x-y\equiv 1,3 \pmod 4$$ From this you may find out $(x,y)$.
