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 } $ and $ s=2 $
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 } $ and $ s=2 $
Now note $ \displaystyle { P(n) - 2015 \equiv (-2015) \mod P(n) } $ and $ s=2 $
$ \displaystyle { \Rightarrow {P(n) - 2015}^t \equiv {-2015}^t \mod P(n) } $ and $ s=2 $
$ \displaystyle { P(P(n) - 2015) } $ and $ s=2 $
$ \displaystyle { \equiv a_k (P(n) - 2015)^k + a_{k-1} (P(n) - 2015)^{k-1} + ... + a_1 (P(n) - 2015)) + a_0 } $ and $ s=2 $
$ \displaystyle { \equiv a_k (- 2015)^k + a_{k-1} (- 2015)^{k-1} + ... + a_1 (- 2015) + a_0 } $ and $ s=2 $
$ \displaystyle { \equiv P(-2015)\mod P(n) } $ and $ s=2 $
But it is given that $ \displaystyle { P(P(n)-2015) \equiv 0 \mod P(n) } $ and $ s=2 $ for all n.
Hence $ \displaystyle { P(-2015) \equiv 0 \mod P(n) } $ and $ s=2 $ 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.
