Pythagorean Triples Problem

Moderator note:

Why couldn't $131$ be the largest of these three numbers? That is, why can't $131$ be the hypotenuse of a right triangle with sides $n$ and $n+1$ ?

You need to explain why $131$ cannot be the largest side of a Pythagoras Triplet, $(n,n+1,131)$ .

I'm under the assumption that $n$ must be an integer, else there would be more than 1 solution.

Here's my working:

Suppose it is possible, then by Pythagoras Theorem, $n^2 + (n+1)^2 = 131^2$ , then $2n^2 + 2n - 17160 = 0$ , or $n^2 + n - 8580 = 0$ . Note that the discriminant of the quadratic equation above is $B^2- 4AC = 34321$ which isn't a perfect square, thus it can't have integer root. In other words, there can't have integer lengths of a right triangle of lengths $n,n+1,131$ with $131$ as its hypotenuse.

Bonus questions:

1. In the working above, why didn't I solve the quadratic equation, but instead just determine the property of its discriminant? Is it sufficient that it be a perfect square to have integer root(s)? Why or why not?

2. Prove that no right triangles with integer sides have a hypotenuse of length $131$ .