Infinite composite polynomials

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homeschool girl

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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. Remember that if a and b are distinct integers, then P(a) - P(b) is divisible by a - b.


I'm not sure where to start.
 
homeschool girl said:
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. Remember that if a and b are distinct integers, then P(a) - P(b) is divisible by a - b.


I'm not sure where to start.
Click to expand...
You need to start with whatever you have learned; since you haven't told us anything about your context, we can't be sure.

It appears that the last sentence represents a theorem you have seen, since it is not obvious on the surface. Is that true? Then that is one place to start.

You also need to make sure you understand the problem. You called this "composite polynomials", which is misleading; do you understand that it is about the value of the function being a composite integer?

Now please tell us what you have learned that you might use to prove that a number is composite.

If you have no specific ideas now, you might just try playing with a specific example. Can you prove, using the same ideas as the theorem referenced, that there are infinitely many positive integer values of x for which P(x) = x^2 is a composite number?
 
the problem could have something to do with finding roots but I have no clue how it relates.

P(x)=x^2 is always composite because it's divisible by x.
 
Like Dr. Peterson, I'd start by thinking about the clue that you have been given.

Suppose k is an integer greater 2 and P(k) is a prime.

Now consider P(k - 1). It is a positive integer. Is it prime or composite?
 
Under the conditions given, does not P(k - 1) have to be divisible by 2?
 
First you have to prove that one such [imath]n[/imath] exists. If [imath]P(n) = q[/imath] where [imath]q>1[/imath] can you find an [imath]m[/imath] such that [imath]P(m)[/imath] is divisible by [imath]q[/imath] ?
 
blamocur said:
First you have to prove that one such n exists
Click to expand...

If P(0) = 0, then P(x) has factor x

If P(0) != 0, then...

homeschool girl said:
Remember that if a and b are distinct integers, then P(a) - P(b) is divisible by a - b.
Click to expand...

Assuming the above, then let b=0 (copy out the above rule and make the substitution b=0)

Now, can you think of a value for "a" that would prove that P(a) is composite?

Given...
x - y is divisible by z
y is divisible by z
Both of the above, together, imply that x is also divisible by z. Can you see why?

Now think what could be done to prove the remaining part of the suggestion by @blamocur in post#9
 
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