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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.

 Feb 13, 2020
 #1
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0

Rip my math ability cannot solve any of Dragonlords or your problems

 Feb 13, 2020
 #2
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+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.

 Feb 13, 2020
 #3
avatar+29257 
+5

For x2 + 1:   Let x = 2n + 1, where n is any positive integer.   Then  x2 + 1 = (2n + 1)2 + 1 = 4n2 + 2n + 1 + 1 = 2(2n2 + 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.

 Feb 13, 2020
edited by Alan  Feb 13, 2020
 #4
avatar+287 
+1

Wow, interesting strategies!

 Feb 17, 2020

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