Question 1160694: 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: Remember that if $a$ and $b$ are distinct integers, then $P(a) - P(b)$ is divisible by $a - b.$
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Let's consider a quadratic polynomial of the form
P(x) = ax^2 + bx + c
where a,b,c are nonnegative integers
If we plug in x = n, then we get
P(x) = ax^2 + bx + c
P(n) = a*n^2 + b*n + c
which isn't much of a change at all.
Note how we have a common 'n' we can factor out from the first two terms, but that third term c does not have an 'n'. At this point, factorization cannot happen.
If we replaced n with something that had c in it, then we can factor. Let n = k*c for some integer k.
We then can say,
P(n) = a*n^2 + b*n + c
P(n) = a*(k*c)^2 + b*(k*c) + c
P(n) = a*k^2*c^2 + b*k*c + c
P(n) = a*k^2*c*c + c*b*k + c*1
P(n) = c*a*k^2*c + c*b*k + c*1
P(n) = c*(a*k^2 + b*k + 1)
showing that P(n) is composite because c is a factor of P(n).
-----------------------------------------------------
To recap: If n = k*c, then P(n) = c*(a*k^2 + b*k + 1) is composite.
This idea can be extended to polynomials of fourth degree, fifth degree, etc. Because k is any integer, and the set of integers is infinitely large, this means there are infinitely many numbers of the form n = k*c that lead to P(n) being composite.
|
|
|
