1

Let $p(x) = 1+a_1x+a_2x^2+\cdots+a_nx^n$ be a polynomial where $a_1,\ldots,a_n$ are integers, and $a_1 + \cdots + a_n$ is even. Prove that there is no integer x such that $p(x) =0$.

I have started this by trying to examine this case when it is only $1+a_1x$ but since $a_1$ cannot be $1$ (it has to add evenly), this will not work without the x value being a fraction.

When you try to examine the next value of $1+a_1x+a_2x^2$ you get the same problem.

How would I start this problem?

Empty
  • 13,012
user2574069
  • 51
  • If there is some $x$ such that $1+a_1x+a^2x_2+\cdots+a_nx^n=0,$ then $x$ divides $1$. There are only two possibilities: $x=\pm1$. –  Apr 19 '14 at 00:48
  • What is the restriction on $n$? If $n = 1$ then $x + 1$ certainly has an integral root. – MT_ Apr 19 '14 at 02:39

4 Answers4

7

Hint: $x$, $x^2$, ..., $x^n$ are either all even or all odd.

Robert Israel
  • 448,999
5

We can prove a stronger result that if $P(0) \equiv P(1) \equiv 1 \pmod 2$ with $P$ having integral coeffs, then $P$ has no integral roots.

Hint: $a, 1-a$ have different parity. Plug them into $P$ and suppose $a$ is an integral root.

Sandeep Silwal
  • 8,135
4

Hint $\ $ Apply the following simple test.

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
0

A different hint:

$a-b \mid p(a) -p(b) $

Calvin Lin
  • 68,864