**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:

**Paper:**RMO 2015 Mumbai**What is this topic:**Polynomial**What are some of the associated concepts:**Modular Arithmetic**Where can learn these topics:**Cheenta**Book Suggestions:**Polynomial by Barbeau