How to generate factors given the b-value of the quadratics is the hypotenuse of a Pythagorean Triple?

Question

My friend has written a nonconstructive proof of the following:

Given two quadratics with integer coefficients,

$$y = x^2 + bx + c, \text{ and } y = x^2 + bx - c,$$

If both quadratics factor to integer roots, then $b$ is the hypotenuse of a Pythagorean Triple.

His proof is thus:

If the roots are integers, then the discriminants of both must perfect squares, i.e. for integers $n,m$, $b^2 -4c = n^2$ and $b^2 +4c = m^2$

Adding the two lines, we get $2b^2 = m^2 + n^2$, where $m^2 + n^2$ is an even hypotenuse of a Pythagorean Triple (by Euclid's method of generating right triangles). (It is even since both discriminants are either both even or both odd.)

So, $2b^2 = m^2 + n^2 = 2h^2$ for some P.T. hypotenuse $h$, such that $b^2 = h^2$

So now I'm curious about the following,

1. Is the converse of this proof also true?

2. And if so, given any P.T. hypotenuse, is there a direct construction of all four roots of the above quadratics? E.g. for $b=13$, the roots are $3, 10$, and $-2, 15$. Can we generate those roots given only the Pythagorean Triple $(5,12,13)$?

3. In general, what is the pattern to the roots?

Note

One interesting thing I've found is that, if $(x-p_1)(x-p_2)$ are factors of the first and $(x-q_1)(x-q_2)$ are factors of the second quadratic, then $b = \frac{p^2 +q^2}{p+q}$ for all permutations of $p$ and $q$.

Answer 1

Let $p_{\pm}(x)=x^2+bx\pm c\in\mathbb Z[x]$, with discriminant $b^2\mp4c$. As you observed, if $p_{\pm}$ have integer roots, then for integers $m,n$ we have $b^2+4c=m^2,b^2-4c=n^2$ so that $m^2+n^2=2b^2$. This does not prove that $b$ is the hypotenuse in a pythagorean triple. For example, the degenerate case $m=n=b$ always solves the equation $m^2+n^2=2b^2$, but obviously not every positive integer is the hypotenuse in a pythagorean triple.

Answer 2

COMMENT.- The equation $X^2+Y^2=2Z^2$ has as general solution the parameterization coming from the identity of two parameters $s,t$ $$(s^2-t^2+2st)^2+(s^2-t^2-2st)^2=2(s^2+t^2)^2$$ Then if you have $2b^2=m^2+n^2$ you should have for some values of $s,t$ $$b=s^2+t^2\\m=s^2-t^2+2st\\n=s^2-t^2-2st$$ Thus, always $b$ is the hypotenuse of a right triangle of legs $s^2-t^2$ and $2st$

On the other hand, the equation $X^2+Y^2=Z^2+W^2$ has also an infinity of solutions then sometimes the right triangle above is not necessarily unique.

With straightforward calculation you can get $c=st(s^2-t^2)$ to form your equations $$y = x^2 + bx + c, \text{ and } y = x^2 + bx - c$$ and $$b^2+4c=m^2\\b^2-4c=n^2$$