If $a$ is an odd integer then $x^2+x-a = 0$ has no integer solutions

Question

I'm suppose to prove by contrapositive that if $a$ is an odd integer then the equation $x^2+x-a=0$ has no integer solution.

By contrapositive:

If the equation $x^2+x - a = 0$ has an integer solution then $a$ is an even integer. So I attempt to apply the quadratic formula and have this result $\frac{-1 \pm \sqrt{1 - 4a}} 2$. I have no idea how I'm suppose to get an integer solution from this, let alone an even solution. I've tried to multiply by the conjugate but it gets really messy and I feel that I'm over-thinking it. Thanks for your help.

Answer 1

Hint: If $x$ is any integer, then $x^2+x$ is even.

Answer 2

$a=2n+1 \to x = \dfrac{-1 \pm \sqrt{1+4(2n+1)}}{2}$.

Problem is to show $8n+5$ can't be a square. Consider $(8k+n)^2 \pmod 8$, as $n$ runs from $0 \to 7$, you can't have it equal $5 \pmod 8$.

Answer 3

$\sqrt{1+4a}$ should be odd to have an integer solution. So, $1+4a=(2k+1)^2$ for some integer $k$. Hence, $a=k(k+1)$ which implies $a$ is even.

Answer 4

Hint $\ $ Show that $\rm\,f(x)\,$ is odd so $\ne 0\,$ whether $\rm\,x\,$ is even or odd. More generally we have

Parity Root Test $\ $ A polynomial $\rm\,f(x)\,$ with integer coefficients has no integer roots when its constant coefficient and coefficient sum are both odd.

Proof $\ $ The test verifies that $\rm\ f(0) \equiv 1\equiv f(1)\ \ (mod\ 2),\ $ i.e. that $\rm\:f(x)\:$ has no roots modulo $2$, hence no integer roots. $\ $ QED

The Parity Root Test generalizes to any ring with a sense of parity, e.g. the Gaussian integers $\rm\: a + b\,{\it i}\ $ for integers $\rm\:a,b.\:$ For much further discussion see this post and also these related posts.