Does it possible to show that the Diophantine equation $X^2-Y^2=N$ has no solution except trivial?

Question

Does it possible to show that the Diophantine equation $X^2-Y^2=N$ (N - odd)has no non-trivial solution?

Answer 1

Suppose $N=CD$ where $C,D$ are odd and greater than $1$ and $C<D.$

Then setting $X-Y=C,\ X+Y=D$ gives $X=(D+C)/2,\ Y=(D-C)/2.$ Since these are not $(N \pm 1)/2$ the solution is not "trivial" in the sense of your last comment.

Answer 2

Basically, the equation $$x^2-y^2=N$$ is such that the set of integer solutions $(x,y)\in {\mathbb Z}^2$ has a bijection (given in older posts) with the set of integer solutions $(a,b)$ such that $a \pm b\equiv N (\mbox{mod }2)$ and $$ab=N.$$

Thus, a natural number $N>0$ has no nontrivial solutions if and only if $N$ is a prime number (odd case) or a number $\equiv 2 (\mbox{mod }4)$.

I am merely rephrasing what's been done above, i.e. clearly $a=x-y$, and $b=x+y$.