If $a$ is an odd integer then $x^2+x-a = 0$ has no integer solutions 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. 
 A: Hint: If $x$ is any integer, then $x^2+x$ is even.
A: 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.
A: $\sqrt{1+4a}$ 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.
A: $a=2n+1 \to x = \dfrac{-1 \pm \sqrt{1+4(2n+1)}}{2}$.
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 $5 \pmod 8$. 
