+24 Idek what to do with the problem: 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.
+1768 Let P(x) = x^2. Then P(1) = 1, p(2) = 4, p(3) = 9, so p(n) is always a perfect square. You can do this for any polynomial that factors, like P(x) = x^3 and P(x) = x(x + 1).
+964 By the prime number theorem, the nth prime is like n*log n. This is not a polynomial, so there must be an n such that p(n) is composite.
+1537 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.
