+305 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.
Hint that I got: Remember that if \(a\) and \(b\) are distinct integers, then \(P(a)-P(b)\) is divisible by \(a-b.\)
Thanks!!! <3
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.
+305 Thank you so much for your response! Guest. I have contacted my friend and got a much more complex solution, your way may work as well, though! :D
