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

$x^4+4$ 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.

• 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 $x > 1$. Commented Jul 25, 2011 at 23:29
• @algebra_fan thanks. this was just silly. Commented Jul 25, 2011 at 23:36
• I've closed the question as a duplicate. Commented Jul 26, 2011 at 0:39
• @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. Commented Jul 26, 2011 at 0:46
• Moreover, Americo's answer on the other question explicitly addresses this inequality. Commented Jul 26, 2011 at 1:00

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 $x \gt 1$.

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:

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

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 $1.$