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$x^4+4$ is composite for $x>1$

I know the Sophie Germain indentity and the get the factorization $$x^4+4 = (x^2+2-2x)(x^2+2+2x)$$

But I am stuck here. I cannot see any general factor here.

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    I am not sure what the question is here. You've written $x^4+4$ as a product of two integers both of which are $ > 1$, so $x^4+4$ is composite for $x > 1$.2011-07-25
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    @algebra_fan thanks. this was just silly.2011-07-25
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    I've closed the question as a duplicate.2011-07-26
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    @Zev It's *not* a dup since the OP already knows the factorization. Given that, it's really a question about proving an inequality - unlike the proposed dup. Could you please reopen it. Please be careful wielding the great power of binding close votes.2011-07-26
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    Very well, I've reopened (out of curiosity, how did you manage to post your answer after the question was closed? My impression was that a banner should appear saying that no new answers would be accepted, or some such thing).2011-07-26
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    Well... if anyone should be able to override closed status, it should be Bill!2011-07-26
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    @Bill: I received at least one flag as a duplicate; I also believe that this could very well go as a *comment* on one of the answers of the other question, asking for clarification - the notion that this is a duplicate is not so unreasonable. However, I do not feel very strongly about it, so I am happy to reopen at your request.2011-07-26
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    Moreover, [Americo's answer](http://math.stackexchange.com/questions/21146/prove-a-number-is-composite/21161#21161) on the other question explicitly addresses this inequality.2011-07-26
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    Related to, if not a duplicate of, http://math.stackexchange.com/questions/21146/prove-a-number-is-composite @Bill, the OP does not seem to be aware of that question.2011-07-26
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    @Zev I see no such *proof* in the answer you cite. Rather, the key inequality is simply stated as a fact there. Look closer.2011-07-26
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    @Zev I agree that, in theory, it should be a duplicate. I don't recall any discussion of issues like this on meta so perhaps it would be worthwhile, i.e. what to do with questions that occur in the solution of another question, but which haven't been (completely) addressed in answers to the other question. It's conceivable that, in some cases, very interesting questions could arise that deserve to get front-page exposure as a new question.2011-07-26
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    "Sophie Germain's identity"?? Googling, I found many references to this phrase, such as http://www.math.ca/crux/v26/n7/page417-428.pdf where the following incorrect statement appears: "As the name says, Sophie Germain's identity was first discovered by Sophie Germain." Shouldn't this be named for Nicholas Bernoulli (1687-1759)? In a 1702 paper Leibniz thought that $x^4 + a^4$ for $a$ real could not be factored into real quadratics, which led Bernoulli to show otherwise in 1719. See pp. 411-412 Kline's "Mathematical Thought From Ancient to Modern Times" (1972).2011-07-26

3 Answers 3

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The factorization that you mention finishes things! For it is clear that if $x\gt 1$, then each of $x^2+2-2x$ and $x^2+2+2x$ is greater than $1$, so $x^4+4$ has a non-trivial integer factor.

For a proof that, for example, $x^2+2-2x \gt 1$ if $x \gt 1$, note that $$x^2+2-2x =(x-1)^2+1$$ and if $x \gt 1$, then $(x-1)^2 \ge 1$.

It is true that there is no constant $k \gt 1$ such that $k$ divides $x^4+4$ for every integer $x \gt 1$, but we don't need that, all we need to show is that there is a non-trivial factor for any $x \gt 1$.

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HINT $\ \ x\:(x+2)+2\ \ge\ x\:(x-2)+2\ \ge\ 2\ $ for $\ x\ge 2$

You may find of interest that this is a special case of a class of cyclotomic factorizations due to Aurifeuille, Le Lasseur and Lucas, the so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\ D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations:

$$\begin{array}{rl} x^4 + 2^2 \quad=& (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\ \frac{x^6 + 3^3}{x^2 + 3} \quad=& (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\ \frac{x^{10} - 5^5}{x^2 - 5} \quad=& (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\ \frac{x^{12} + 6^6}{x^4 + 36} \quad=& (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\ \end{array}$$

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A version in two variables is $$ ( x^2 + 2 x y + 2 y^2) (x^2 - 2 x y + 2 y^2) = x^4 + 4 y^4 $$ where both quadratic forms are positive definite, so that there are only a finite number of $(x,y)$ pairs such that either factor is $1.$