2

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.

kingW3
  • 13,496
ponderingdev
  • 558

4 Answers4

9

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

Berci
  • 90,745
  • wow...okay it makes sense now. i'd have to make two cases right? one when it's odd and one when it's even. do you know that property just from practice or is there any special intuition behind that? I don't think I would have figured that out by myself to be honest – ponderingdev Feb 16 '15 at 20:34
  • @curiousmind1995: $x^2+x=x(x+1)$ which is a product of 2 consecutive integers, so one would be even and so is the product. – Extremal Feb 16 '15 at 20:45
  • Makes perfect sense. Thank you! – ponderingdev Feb 16 '15 at 20:47
  • My 'special intuition' behind was that $x^2$ has the same parity as $x$, and adding two integers of the same parity gives always even. – Berci Feb 17 '15 at 20:54
2

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.

Community
  • 1
Bill Dubuque
  • 272,048
1

$\sqrt{1+4a}$ 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.

Extremal
  • 5,785
0

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

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 $5 \pmod 8$.

DeepSea
  • 77,651