Question 1160694
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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) = <font color=blue size=4>c</font>*a*k^2*c + <font color=blue size=4>c</font>*b*k + <font color=blue size=4>c</font>*1
P(n) = <font color=blue size=4>c</font>*(a*k^2 + b*k + 1)
showing that P(n) is composite because c is a factor of P(n). 


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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.
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