0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Your title asks for a proof of a false claim. You can surely think of a reasonable title... $\endgroup$ – Mariano Suárez-Álvarez Nov 6 '16 at 7:01
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $p(22) - p(2) = 2q - q = q$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.