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

 A: 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. 
A: 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 $a,b$.
A: 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
$$p^2+2apq+2bq^2=0$$
This implies $p$ is even.  So writing $p=2p'$ and then dividing through by $2$, we have
$$2p'^2+2ap'q+bq^2=0$$
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.)
A: 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.
