# Prove there does not exist a polynomial

Can anyone help me solve this? I have no idea about how to tackle this problem.

Prove that there does not exist any polynomial $p(x) \in \mathbb{Z}[x]$ such that $p(2) = q$ and $p(22) = 2q$, where $q$ is an odd prime.

• Your title asks for a proof of a false claim. You can surely think of a reasonable title... – Mariano Suárez-Álvarez Nov 6 '16 at 7:01

Hint

Suppose there is a polynomial $p(x)=a_nx^n+ \dotsb + a_1x+a_0$ (where $a_i \in \mathbb{Z}$) that satisfies the given conditions. Then $$p(2)=q \implies a_n(2^n)+ \dotsb + a_1(2)+a_0=q.$$ But $q$ is given to be odd, therefore $a_0$ must be odd.

Now try to use the other condition to see if you can arrive at some sort of contradiction.

Hint: $p(a) - p(b)$ is divisible by $a-b$.

Notice for any integer $k > 0$, we have the factorization

$$x^k - y^k = (x-y)(x^{k-1} + x^{k-2}y + \cdots + xy^{k-2} + y^{k-1})$$

and $x - y$ is a factor of $x^k - y^k$. This implies for any $f(x) = \sum\limits_{k=0}^n a_k x^k \in \mathbb{Z}[x]$, we have

$$f(x) - f(y) = (x-y)\left(\sum_{k=1}^n a_k \sum_{\ell=0}^{k-1} x^{k-1-\ell}y^\ell\right)$$ Since the last factor is a polynomial in $x,y$ with integer coefficients, it will be an integer whenever $x,y$ are integers. This leads to a useful lemma:

For any integer $u, v$ and polynomial $f(x)$ with integer coefficients, $u - v$ divides $f(u) - f(v)$.

Apply this to the case $f(x) = p(x)$, $u = 22$ and $v = 2$. If $q$ is an odd prime, this will lead to a contradiction that $20 = 22 - 2$ divides $p(22) - p(2) = 2q - q = q$.