document.write( "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.
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document.write( "hint: Remember that if $a$ and $b$ are distinct integers, then $P(a) - P(b)$ is divisible by $a - b.$ \n" );
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Algebra.Com's Answer #784093 by jim_thompson5910(35256)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Let's consider a quadratic polynomial of the form \n" ); document.write( "P(x) = ax^2 + bx + c \n" ); document.write( "where a,b,c are nonnegative integers\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If we plug in x = n, then we get \n" ); document.write( "P(x) = ax^2 + bx + c \n" ); document.write( "P(n) = a*n^2 + b*n + c \n" ); document.write( "which isn't much of a change at all. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "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. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If we replaced n with something that had c in it, then we can factor. Let n = k*c for some integer k. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We then can say, \n" ); document.write( "P(n) = a*n^2 + b*n + c \n" ); document.write( "P(n) = a*(k*c)^2 + b*(k*c) + c \n" ); document.write( "P(n) = a*k^2*c^2 + b*k*c + c \n" ); document.write( "P(n) = a*k^2*c*c + c*b*k + c*1 \n" ); document.write( "P(n) = c*a*k^2*c + c*b*k + c*1 \n" ); document.write( "P(n) = c*(a*k^2 + b*k + 1) \n" ); document.write( "showing that P(n) is composite because c is a factor of P(n). \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "-----------------------------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "To recap: If n = k*c, then P(n) = c*(a*k^2 + b*k + 1) is composite.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "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. \n" ); document.write( " \n" ); document.write( " |