A number theory problem by Sayantan De

A square integer is congruent to 0 or 1 modulo 4. Therefore, $x^2-y^2\equiv0,1$ or $3\pmod{4}$ . Since $50\equiv 2\pmod{4}$ , there exist no integers $x,y$ so that $x^2-y^2=50$ .

We are interested in determining $x,y \in \mathbb{N}$ such that $x^2-y^2 = 50$ . Taking $(x+y)(x-y) = 2^{1}5^{2}$ , we will have six positive integral divisors of $50 \Rightarrow d(50) = 1,2,5,10,25,50.$ If we take the following factor combinations:

$x+y = 10, 25, 50$ AND $x-y=5,2,1$

we respectively find that $(x,y) = (\frac{15}{2},\frac{5}{2}); (\frac{27}{2},\frac{23}{2}); (\frac{51}{2},\frac{49}{2}) \notin \mathbb{N}.$ Therefore $x^2-y^2 = 50$ is NOT possible for $x,y \in \mathbb{N}.$