Proof that polynomial with given rules is not possible

Question

Prove that there is no polynomial $P(x)$ with whole number coefficients for which: $$P(7)=5\\P(15)=9$$

So what I know, a polynomial is defined as:

$P(x)=a_0+a_1x+a_2x^2+...+a_nx^n$

and we need $a_0,a_1,...,a_n\in\Bbb{Z}$

so first I wrote down:

$P(7):a_0+7a_1+7^2a_2+...+7^na_n=5$

thought of simplyfing this somehow to make it usable in any way

$7(a_1+7a_2+...+7^{n-1}a_n)+a_0=5\\7(a_1+7(a_2+7(a_3+...+7a_n)))+a_0=5$

I thought now of maybe trying to move something to the other side

$$a_1+7(a_2+...+7a_n)={5-a_0\over7}$$

I could go and do this forever now and I'm not sure how to connect that with proving that $P(x)$ with the given rules doesn't exist. Stuck here.

Answer 1

Hint:

If $p\in \mathbb{Z}[x]$ then for every integers $x,a$ we have $$x-a\mid p(x)-p(a)$$

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This fact is easy to prove. If we divide $p(x)$ with $x-a$ we get remainder to be a constant say $c$ so we have $$p(x) = k(x)(x-a)+c\;\;\;\;(*)$$ Now, what is $c$? Set in $(*)$ $x=a$ and we get $c= p(a)$ and we are done.