+0  
 
0
100
2
avatar+278 

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

 Feb 11, 2020
 #1
avatar
0

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.

 Feb 13, 2020
 #2
avatar+278 
0

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

 Feb 13, 2020

38 Online Users

avatar
avatar
avatar
avatar
avatar
avatar