Let \(P(x)\) be a nonconstant polynomial, where all the coefficients are nonnegative integers. Prove that there exist infinitely many positive integers \(n\) such that \(P(n)\) is composite.

I can prove it for all cases that the constant is 0 and all real numbers OTHER than 1.

I don't know how to prove that it is composite for \(P(x)=x^{2}+1\)

Any help will be appreciated.

EpicWater Feb 13, 2020

#2**+1 **

We know that P(a) - P(b) is divisible by a - b. This means if a - b is not a prime number, then P(a) - P(b) is not a prime number. So by modular arithmetic, we can find a positive integer n such that P(n) is composite.

If P(a_1) and P(a_2) are composite, then P(a_1) - P(a_2) will be divisible by some prime number. Similarly, if P(a_1) - P(a_2) is divisible by some prime number, and P(a_1) or P(a_2) is composite, then the other one is also composite.

We can then take a_1 = n. So there exists a positive integer a_2 such that P(a_2) is composite. But since P(a_2) is composite, we can apply the result above, to find a positive integer a_3 such that P(a_3) is composite. In this way, we can generate an infinite number of positive integers n such that P(n) is composite.

Guest Feb 13, 2020

#3**+5 **

For x^{2} + 1: Let x = 2n + 1, where n is any positive integer. Then x^{2} + 1 = (2n + 1)^{2} + 1 = 4n^{2} + 2n + 1 + 1 = 2(2n^{2} + n + 1) which is even, and hence composite (since x starts at 3 when n = 1). There are therefore an infinite number of positive integers (all the odd numbers > 1) that satisfy the condition.

Alan Feb 13, 2020