Prime @ perfect square

Question

Given that x, y fulfilled the following conditions. I, x - y is a prime number. II, xy is a perfect square number. III, x » 2015. Calculate the smallest possible value of x

Answer 1

If $n^2=xy$ with $x-y$ prime, then the greatest common divisor of $x $ and $y$ must be a divisor of $x-y$, so must be $1$ or $x-y$. If the gcd is $1$, the fundamental theorem of arithmetic then tells you both $x$ and $y$ must be perfect squares. The difference between two perfect squares $x$ and $y$ is prime only for $x=(n+1)^2$, $y=n^2$ (see this question). Now all you have to do is find the smallest $n$ such that $2n+1$ is prime and $x=(n+1)^2\ge 2015$.

If the gcd is $x-y=p$ and is prime, then again by the fundamental theorem of arithmetic $\frac x p$ and $\frac y p$ must be perfect squares. So $p=x-y=A^2p-B^2p=p(A^2-B^2)$ for some positive integers $A$ and $B$, so $A^2-B^2=1$. This is impossible.

Answer 2

$x-y=p$ so $x=y+p$ and $xy=y(y+p)=n^2$.

$p\ne 2$ because then $x=a+1$ and $y=a-1$ and $xy=a^2-1\ne n^2$. So $p$ is odd and $x,y$ have different parities.

If $x=bj, y=bk$ then $x-y=b(j-k)$ is composite; so $\gcd{x,y}=1$. Hence both $y$ and $(y+p)$ must be squares.

So look at the first square larger than $2015$. $45^2=2025$. Try $(y+p)=x=2025$ recalling that $y$ itself is a square. Try $y=44^2=1936$. As it turns out $2025-1936=89$ and $89$ is prime.

So $x=2025, y=1936$ gives the smallest possible value of $x$ that satisfies the conditions.