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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.

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  • $\begingroup$ Can you kindly tell me how to use it...so that I can incorporate $\endgroup$ – user854451 Nov 28 '20 at 10:58
  • $\begingroup$ 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 $\endgroup$ – user854451 Nov 28 '20 at 11:10
  • $\begingroup$ Nikhil why did you delete your post...it was beneficial $\endgroup$ – user854451 Nov 28 '20 at 12:23
  • $\begingroup$ Whatever reuns said was correct but I feel too abrupt....a little more explanation can help.me understand better.. $\endgroup$ – user854451 Nov 28 '20 at 14:32
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If a prime $p$ divides $f(2^{mp})$ then it divides $f(2^m)$.

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  • $\begingroup$ Reuns...can u explain a little bit $\endgroup$ – user854451 Nov 28 '20 at 15:27
  • $\begingroup$ $2^p \equiv 2\bmod p$ thus $f((2^p)^m) \equiv f(2^m)\bmod p$ $\endgroup$ – reuns Nov 28 '20 at 15:49
  • $\begingroup$ Fermat's little theorem $\endgroup$ – iBug Nov 28 '20 at 18:57
  • $\begingroup$ What's the next step after this? $\endgroup$ – cansomeonehelpmeout Nov 29 '20 at 9:24
  • $\begingroup$ @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. $\endgroup$ – reuns Nov 29 '20 at 12:59

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