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

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 $$q$$.

• You claim $b$ is a hypotenuse, but you prove $b^2$ is a hypotenuse. Mar 8, 2020 at 22:08
• I am confused a little. The hypotenuse squared of a right angle triangle with $m$ and $n$ as sides is $m^2+n^2$. But you have $2b^2=m^2+n^2$. Then $b$ can't be a perfect square. Mar 8, 2020 at 22:12
• Ngo Gol: there is absolutely no reason for all the vertical space in your original post. Mar 8, 2020 at 22:13
• Editted. $b$ is definitely a hypotenuse, I apologize that the proof is clunky. Mar 8, 2020 at 22:26

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.
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 $$b^2+4c=m^2\\b^2-4c=n^2$$