# To show P is a zero polynomial

Suppose that $$P$$ is a polynomial with integer coefficients that $$n$$ divides $$P(2^n)$$ for every positive integer $$n$$.

Prove that $$P$$ must be the zero polynomial.

What I did was apply some induction on the expression by considering $$P (x)= a_nx^n+ \cdots +a_0$$ which results in nothing for proving the required result. Any hints/solution would be appreciated.

• Can you kindly tell me how to use it...so that I can incorporate – user854451 Nov 28 '20 at 10:58
• Nikhil.....it would be highly beneficial if you can provide a rigorous proof...in the answer sectiom instead od the comments as it is not so helpful for me to catch – user854451 Nov 28 '20 at 11:10
• Nikhil why did you delete your post...it was beneficial – user854451 Nov 28 '20 at 12:23
• Whatever reuns said was correct but I feel too abrupt....a little more explanation can help.me understand better.. – user854451 Nov 28 '20 at 14:32

If a prime $$p$$ divides $$f(2^{mp})$$ then it divides $$f(2^m)$$.
• $2^p \equiv 2\bmod p$ thus $f((2^p)^m) \equiv f(2^m)\bmod p$ – reuns Nov 28 '20 at 15:49
• @cansomeonehelpmeout If $P\ne 0$ then for $m$ large enough $P(2^m)\ne 0$ so that every prime divides $P(2^{mp})$ thus $P(2^m)$ is a contradiction. – reuns Nov 29 '20 at 12:59