A number theory problem by v a

let $x=y+b$ , thus $y^2+2yb+b^2-y^2=31$ , simplifying and factoring gives $b(2y+b)=31$ .

Now $31$ is prime, so the only way of making 31 by multplying two numbers would be with $1 \times 31$ . Obviously, $2y+b$ must be greater than $b$ , so we let $b = 1$ and $2y+b = 31$ . by substitution, $2y+1 = 31$ , which means $y = 15$ , and $x = 15+1 = 16$ .

Because the only factors of 31 are 1 and 31, there is only the one solution, $16^2-15^2=31$

See, now we have 31, which is a prime. So the only way is 1 x 31. So we have $x+y = 31$ . $x-y=1$ . Adding, we have $2x = 32$ and so $x = 16$ . From the above equations, $x+y=31$ Substituting $x=16$ , we have $16+y = 31$ . which implies $y = 31 -16 = 15$ . So there is only one solution.