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Show that $n^{4}-20n^{2}+4$ is a composite whenever $n$ is an integer. I'm having trouble putting together a proof for this problem.

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Hint: Did you try factoring the expression? Regards – Amzoti Feb 10 '13 at 0:49

You can factor this as $(n^2-4n-2)(n^2+4n-2)$. Now you just have to show that neither factor is $1$.

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Is it necessary to show that neither factor equals one to prove that the function is a composite whenever n is an integer? – user61646 Feb 10 '13 at 18:04
@user61646: Yes. If either factor could be $\pm 1$, it might be that the other is prime and the whole product could be prime. For example, I could argue that $m^2-1$ is never prime because of the factorization $(m+1)(m-1)$. This fails for $m=2$. But it is not hard. Note that $n^2 \pm 4n -2=(n \pm 2)^2-6$ and $5,7$ are not squares. – Ross Millikan Feb 10 '13 at 18:13

I wanted to explain where the factorization in Ross Milikan's answer comes from, as it may not be evident at first sight.

Looking at $n^4-20n^2+4$, it doesn't seem like the quadratic $x^2-20x+4$ factors nicely. However, it looks fairly close to $x^2-4x+4$ which we know is $(x-2)^2$. This leads us to write $$n^4-20n^2+4=\left(n^4-4n^2+4\right)-\left(16n^2\right).$$ This becomes $$\left(n^2-2\right)^2 -(4n)^2,$$ which is a difference of squares, and factors into $$(n^2-4n-2)(n^2+4n-2).$$

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+1 Nice work, Eric! – amWhy Feb 10 '13 at 2:07

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