Prove: For odd integers $a$ and $b$, the equation $x^2 + 2 a x + 2 b = 0$ has no integer or rational roots.

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.

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 $a,b$.