If $a$ and $b$ are odd integers, prove that the equation

$$x^2 + 2ax + 2b = 0$$

has no integer or rational roots.

  • 3
    $\begingroup$ You cannot use $m+1$ and $n+1$ for the odd integers. If $m$ and/or $n$ are chosen odd, then your expressions become even. You need to set $a=2m+1$ and $b=2n+1$ and then use the discriminant. Verify that the discriminant becomes $16m^2+16m-16n-4$. It then becomes the challenge to prove that this expression does not produce perfect squares... $\endgroup$ – imranfat Sep 12 '16 at 2:29
  • $\begingroup$ Sorry... that's what I meant. $\endgroup$ – user193203821309 Sep 12 '16 at 2:48
  • $\begingroup$ @differentialequation Since the question does not reflect what you mean, you should edit the question. $\endgroup$ – Erick Wong Sep 12 '16 at 3:02
  • 1
    $\begingroup$ Do not radically alter the question after it has gotten answers, please. We don't want that work go to waste. This rule is community enforced in several ways. Experienced users (with the privilege) rolling back such edits is one of the more efficient ones :-) $\endgroup$ – Jyrki Lahtonen Sep 16 '16 at 19:45

The discriminant is $4a^2 - 8b = 2^2(\sqrt{a^2 - 2b})^2$.

For rational root(s) to exist, $a^2 - 2b$ has to be a perfect square. Let's assume that this is so, i.e. $a^2 - 2b = n^2$ where $n$ is an integer.

Rearrange to get $a^2 - n^2 = 2b\implies (a+n)(a-n) = 2b$

Note that $a+n$ and $a-n$ have the same parity. Since the RHS is even, both $a+n$ and $a-n$ are even. But then their product would be a multiple of $4$, which means that $b$ would also be even. This is a violation of the initial conditions.

Since we've arrived at a contradiction, the equation cannot have rational roots.

  • $\begingroup$ A nice approach. $\endgroup$ – imranfat Sep 12 '16 at 3:07
  • $\begingroup$ @imranfat Thanks. $\endgroup$ – Deepak Sep 12 '16 at 3:09

By the rational root theorem, any factorisation of the given polynomial must be of the form $(x+c)(x+d)$, where $cd$ is a factorisation of $2b$ into two integers. We have $c+d=2a$.

The prime factors of $2b$ must now be distributed between $c$ and $d$; there is only one even prime factor (2) and the rest are odd. Without loss of generality, assign the factor of 2 to $c$. No matter how the remaining odd factors are distributed, $c$ must remain even because of the 2 and $d$ must remain odd because there is no factor of 2 in it at all. Hence $c+d$ must be odd, which contradicts their sum $2a$ being even.

Therefore $x^2+2ax+2b=0$ has no integer – or rational – roots for all odd integers $a,b$.

  • 2
    $\begingroup$ Elegant! Very nice intuition $\endgroup$ – Vladimir Vargas Sep 12 '16 at 2:46

In a word: Eisenstein.

In a few more words, if $x^2+2ax+2b=0$ had a rational root, say $x=p/q$ with $\gcd(p,q)=1$, then, on multiplying through by $q^2$, we have


This implies $p$ is even. So writing $p=2p'$ and then dividing through by $2$, we have


which implies $q$ is even (since $b$ is odd). But that contradicts $\gcd(p,q)=1$ (i.e., the fact we can write any fraction in reduced form). Thus $x^2+2ax+b=0$ cannot have a rational root if $b$ is odd. (Note, the assumption that $a$ is also odd is irrelevant to the proof.)

  • $\begingroup$ This method is good. You can also prove $q=1$, as $q | 0=p^2+2apq+2bq^2$ implies $q | p^2$, but $gcd(p, q)=1$. $\endgroup$ – S. Y Sep 16 '16 at 14:25
  • $\begingroup$ @S.Y, nice alternative. $\endgroup$ – Barry Cipra Sep 16 '16 at 15:01

The square root of the discriminant of the roots is

$ 2 \sqrt { a^2 -2b} $

This is never a perfect square as it gives 3 mod 4, which is not a quadratic residue.

  • $\begingroup$ @Deepak thanks. Corrected $\endgroup$ – N.S.JOHN Sep 12 '16 at 3:08
  • $\begingroup$ That should be $b$ not $b^2$. $\endgroup$ – Deepak Sep 17 '16 at 1:52

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