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There is this problem that I tried but there are still some questions, confusions and doubts.

There exist infinitely many natural numbers $a$, such that for any natural number $n$, the number $z=n^4 + a$ is a composite number.

How I did it: Let $a=3n^4$, so $z=n^4+3n^4=(2n^2)^2$, since $2n^2>1$, so $z$ is composite.

First I am not sure whether my approach is correct, because for any $a$, it is not the case that there is any natural number $n$ so that $a=3n^4$. Try $a=1$, then $1=3n^4$, then $n$ cannot be a natural number.

Is there something wrong with my solution? Or is there something wrong with my doubt? If so, how can we then solve the problem?

Many many thanks!

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  • $\begingroup$ This isn't correct: You've shown that for any fixed $n$, there is a number $a$ such that $n^4 + a$ is composite. You want to start with an $a$ and show that all numbers of the form $n^4 + a$ are composite, and then make sure there are infinitely many such $a$. $\endgroup$ – user61527 Jan 28 '14 at 0:02
  • $\begingroup$ Your logic is wrong. You are looking for a specific number $a$ (actually, infinitely many of them, but let's just try to find one for a start) such that for every $n$, the number $n^4+a$ is composite. But if you say $a=3n^4$ then this is not a specific number, it is a variable. You need $a=1$ (except that doesn't work) or $a=2$ (except that doen't work either), or. . . $\endgroup$ – David Jan 28 '14 at 0:03
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$\begin{eqnarray}{\bf Hint}\ \ \ n^4+4k^4 &\,=\,& \overbrace{(n^2\!+2k^2)^2}^{\rm\!\!\! complete\ the\ square\!\!\!}-(2nk)^2\ \ \text{so, factoring the} \it\text{ difference of squares}\\ &\,=\,& (n^2\!+2k^2-2nk)\,(n^2\!+2k^2+2nk)\\ &\,=\,&(\underbrace{(n-k)^2}_{\rm\!\!\!\!\!\!\!\!\!\!\! complete\ the\ square\!\!\!\!\!\!\!\!} +\,k^2)\ \ \underbrace{((n+k)^2}_{\rm\!\!\!\!\!\!\!\!\!\!\! complete\ the\ square\!\!\!\!\!\!\!\!} +k^2)\\ \end{eqnarray}$

which is composite for $\,k \ge 2\,$ since both factors have form $\,j^2\!+k^2\ge 4.$

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Hint: $$\eqalign{n^4+a &=(n^2+\sqrt a)^2-(2\sqrt a)n^2\cr &=(n^2+(\sqrt2{\root4\of a})n+\sqrt a)(n^2-(\sqrt2{\root4\of a})n+\sqrt a)\cr}$$ So we have a factorisation of $n^4+a$, only problem is that the coefficients in the factorisation don't look like integers. Can you find special values of $a$ such that all these coefficients are integers?

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$$ a = 4 b^4 $$ $$ n^4 + 4 b^4 = (n^2 - 2 n b + 2 b^2)(n^2 + 2 n b + 2 b^2) $$

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  • $\begingroup$ We have already been through this question (and answer) too much times here... $\endgroup$ – chubakueno Jan 28 '14 at 0:18
  • $\begingroup$ @chubakueno, well, yes. $\endgroup$ – Will Jagy Jan 28 '14 at 0:43

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