0
$\begingroup$

Say you were asked to prove that the polynomial $$n^4 + 4$$ is composite $$\forall(n>1)$$.

The proof basically states that given that $$n^4 + 4 = (n^2 - 2n + 2) * (n^2 + 2n + 2)$$ (easily verifiable via for example long division) and that both of those squared polynomials are greater than 1 for $$n > 1$$ it follows that the original polynomial is composite.

What I am trying to understand is what line of reasoning would lead someone to attempt to factor the original polynomial (rather than say induction) and, more importantly, how one would arrive at one of these two factors (given one finding the other is trivial).

$\endgroup$
  • $\begingroup$ Sophie Germain Identity. $\endgroup$ – user236182 Oct 8 '16 at 19:27
  • $\begingroup$ So is the answer to just know all possible identities off hand? $\endgroup$ – Abraham P Oct 8 '16 at 19:30
  • 1
    $\begingroup$ $$a^4+4b^4=a^4+4a^2b^2+4b^4-4a^2b^2$$ $$=\left(a^2+b^2\right)^2-(2ab)^2$$ $$=\left(a^2+b^2-2ab\right)\left(a^2+b^2+2ab\right)$$ In this case, $$n^4+4=n^4+4n^2+4-4n^2$$ $$=\left(n^2+2\right)^2-(2n)^2$$ $$=\left(n^2+2-2n\right)\left(n^2+2+2n\right)$$ $\endgroup$ – user236182 Oct 8 '16 at 19:32
  • $\begingroup$ The polynomial $n^4 + 4$ can be factored as $$ n^4+4 = \prod_{k=0}^3 (n - z_k). $$ where $z_k = \sqrt{2}\mathrm{e}^{\mathrm{i}(2k+1)\mathrm{\pi}/4}$ for $0\leqslant k\leqslant 3$. As $z_0$ and $z_3$ are conjuguates, same for $z_1$ and $z_2$, we have $$ \begin{align*} n^4 + 4 &= \bigl(n^2 - (z_0 + \overline{z_0}) + z_0\overline{z_0}\bigr) \bigl(n^2 - (z_1 + \overline{z_1}) + z_1\overline{z_1}\bigr) \\ &= (n^2 - 2n + 2)(n^2 + 2n + 2) \end{align*} $$ $\endgroup$ – Éric Guirbal Oct 8 '16 at 20:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.