Let $x^2+mx+n$ and $x^2+mx-n$ give integer roots where $(m,n)$ are integers. Show that $n$ is divisible by $6$

My attempt:

Since the roots are integers then the discriminants of both the equations should be perfect squares.

Let $a=\sqrt{m^2-4n}$ and $b=\sqrt{m^2+4n}$, then $(ab)^2=m^4-16n^2$ where $(a,b,m,n)$ are all integers. I am stuck here...


We have $a^2 + b^2 = 2 m^2$, whose solution in the integers is

$$a = x^2+2xy-y^2, b = y^2 + 2xy - x^2, m = x^2 + y^2, \text { for } x, y \in \mathbb{N}$$

Hence, $ 4n = m^2 - a^2 = 4 xy (x-y) ( x+y)$

It is clear that $ xy (x-y)(x+y)$ must be a multiple of 6.

Proof of classification: (I'm slightly surprised I can't find a derivation on this site, but I'm bad at searching)

If $a$ is even, then clearly $b$ is even, and thus so is $m$, and we can then divide out by 2. Hence, we may assume that $a$ and $b$ are odd, so $ a = 2p-1, b = 2q-1$.

Observe that $m^2 = ( p-q) ^2 + (p+q-1)^2$, hence by the classification of pythagorean triples, we have

$$ m = x^2 + y^2, (p-q) = x^2 - y^2, (p+q-1) = 2xy$$

In other words,

$$a = x^2+2xy-y^2, b = y^2 + 2xy - x^2, m = x^2 + y^2, \text { for } x, y \in \mathbb{N}$$

| cite | improve this answer | |
  • 1
    $\begingroup$ How did you find the solution in integers? $\endgroup$ – Hashir Omer May 10 '14 at 13:19
  • $\begingroup$ @HashirOmer It is a "well known fact", similar to the classification of Pythagorean triples. $\endgroup$ – Calvin Lin May 10 '14 at 13:21
  • $\begingroup$ Its not a well known fact to me. Can you show the working? $\endgroup$ – Hashir Omer May 10 '14 at 13:23
  • $\begingroup$ @HashirOmer Done. $\endgroup$ – Calvin Lin May 10 '14 at 13:37
  • $\begingroup$ @HashirOmer See also Fibonacci's Lost Theorem $\equiv {\rm FLT}_4\ \ $ $\endgroup$ – Bill Dubuque Jun 6 '19 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.