Let $P(x)$ be a polynomial with integer coefficients

Question

Let $P(x)$ be a polynomial with integer coefficients such that $P(2003)\cdot P(2004)=2005$.

How many integer roots does our polynomial have?

I have no idea how where to start on this problem.

Answer 1

Hint $\ P(\color{#c00}{2003})P(\color{#0a0}{2004}) = \color{#c0f}{2005}\,\Rightarrow\, {\rm mod}\ 2\!:\ P(\color{#c00}1) P(\color{#0a0}0) \equiv \color{#c0f}1\,$ so $P(1) \equiv 1 \equiv P(0).\,$ This means that $\,P(x)\,$ has no roots mod $\,2,\,$ hence no integer roots.

Remark $\ $ This is a special case of the following

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.

Answer 2

Hint: Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0$. Show that $P(2003)$ and $P(2004)$ are odd. Use this to show that the sum $\sum a_i$ of the coefficients is odd, and $a_0$ is odd.

Use the above facts to show that $P(k)$ is odd for any integer $k$.