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I've recently started self-studying through Niven's Introduction to the Theory of Numbers and had questions on a few of the problems. In particular, I'm not sure how to show that $n^4+4$ is composite for all $n>1$. I've tried my usual go-to methods of induction and breaking it up into cases by congruence classes. What are some other angles of attack? I always love a new way to take on a problem.

On a side note, to check my work or give me a hint when stuck I couldn't find any solutions for this textbook. (I've seen the "Hints" and "Answers" sections at the end, but comparatively few problems are addressed there.) Can anyone help me find a reference?

Thank you!

Bill Dubuque
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BGreen
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1 Answers1

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In the specific example that you give completing the square yields a difference of squares, therefore it yields a factorization, which is easily proved nontrivial. More generally

$$\begin{eqnarray} \overbrace{\color{#0a0}{n^4+4k^4}}^{\rm incomplete\ \large \Box}\!\! &=\,& \overbrace{\color{#0a0}{(n^2\!+2k^2)^2}}^{\rm\!\!\! completed\ \large \Box\!\!\!}\!\!-\!(\color{#c00}{2nk})^2\ \ \text{so factoring this} \textit{ difference of squares}\\[.2em] &\,=\,& (n^2\!+2k^2\ -\,\ \color{#c00}{2nk})\,(n^2\!+2k^2+\,\color{#c00}{2nk})\\[.2em] &\,=\,&(\underbrace{(n-k)^2}_{\rm\!\!\!\!\!\!\!\!\!\!\! complete\ the\ square\!\!\!\!\!\!\!\!}\ +\ \,k^2)\ \ \underbrace{((n+k)^2}_{\rm\!\!\!\!\!\!\!\!\!\!\! complete\ the\ square\!\!\!\!\!\!\!\!\!\!\!\!\!\!} +\,k^2)\\ \end{eqnarray}$$

which is composite if $\,|k| > 1\,$ or $\,n\neq \pm1,0\,$ since then both factors have form $\,j^2+k^2\ge 2.$

This is sometimes called the Sophie Germain Identity .

Bill Dubuque
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