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December 27, 2015

Polynomial with positive integers | RMO 2015 Mumbai Region)

This is a problem from Regional Mathematics Olympiad, RMO 2015 Mumbai Region based on Polynomial with positive integers. Try to solve it.
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Problem: Let P(x) be a polynomial whose coefficients are positive integers. If P(n) divides P(P(n) -2015) for every natural number n, prove that P(-2015) = 0.

Discussion: 

Let \displaystyle { P(x) = a_k x^k + a_{k-1} x^{k-1} + a_{k-2} x^{k-2} + ... + a_1 x + a_0 }

Then \displaystyle { P(P(n) - 2015) = a_k (P(n) - 2015)^k + a_{k-1} (P(n) - 2015)^{k-1} + ... + a_1 (P(n) - 2015)) + a_0 }

Now note \displaystyle { P(n) - 2015 \equiv (-2015) \mod P(n) }

\displaystyle { \Rightarrow {P(n) - 2015}^t \equiv {-2015}^t \mod P(n) }

\displaystyle { P(P(n) - 2015) }
\displaystyle { \equiv a_k (P(n) - 2015)^k + a_{k-1} (P(n) - 2015)^{k-1} + ... + a_1 (P(n) - 2015)) + a_0 }
\displaystyle { \equiv a_k (- 2015)^k + a_{k-1} (- 2015)^{k-1} + ... + a_1 (- 2015) + a_0 }
\displaystyle { \equiv P(-2015)\mod P(n) }

But it is given that \displaystyle { P(P(n)-2015) \equiv 0 \mod P(n) }  for all n.
Hence \displaystyle { P(-2015) \equiv 0 \mod P(n) }  for all n.

Note that P(-2015) is a fixed number, hence with finitely many divisors.

As a_k is positive, by increasing n arbitrarily, we can increase the value of P(n) infinitely.

But infinitely many numbers cannot divide a finite number (P(-2015)) unless it is equal to 0.

There fore P(-2015) = 0.

Chatuspathi:

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