TITLE: Factor $x^4 + 64$
QUESTION [8 upvotes]: I found a page where this problem was solved but his english is broken so its difficult to understand his explanation. His first step was to divide the constant, 64, by the exponent 4. What is his reasoning behind this step?
http://www.wyzant.com/resources/answers/163/factor_x_4_64

REPLY [2 votes]: The author's reasoning appears to be that it works in this situation so that must be what you have to do. The problem is that it only works for this particular problem. As MJD points out in the comments, Germain's identity shows $a^4+4b^4$ can be factored in the integers. In order to factor this, you need to add and subtract $4a^2b^2$. Then we get $$a^4+4b^4 = a^4+4a^2b^2+4b^4-4a^2b^2 = (a^2+2b^2)-(2ab)^2 = (a^2+2b^2-2ab)(a^2+2b^2+2ab)$$
The only situation where $4b^4\div 4 = 4b^2$ is when $b=2$ and hence $4b^4 = 64$ and $4b^2 = 16$. If you try the same technique if dividing by $4$ on $x^4+4$, which fits the form of Germain's identity with $a=x$ and $b=1$, it fails.