Sums of Fourth Powers While fooling around on my calculator I found:
$$7^4 + 8^4 + (7 + 8)^4 = 2 * 13^4$$
$$11^4 + 24^4 + (11 + 24)^4 = 2 * 31^4$$
I'm intrigued but I can't explain why these two equations are true. Are these coincidences or is there a formula/theorem explaining them?
 A: First, observe $$a^{4}+b^{4}+(a+b)^{4}=2a^{4}+4a^{3}b+6a^{2}b^{2}+4ab^{3}+2b^{4}\\=2(a^{4}+2a^{3}b+3a^{2}b^{2}+2ab^{3}+b^{4})\\=2(a^{2}+ab+b^{2})^{2}$$
Your identities arise when $a^{2}+ab+b^{2}$ is itself a perfect square. Solutions to this equation are generated similarly to pythagorean triples: we paramaterise by coprime integers $m,n$. $$a=m^{2}-n^{2}\\b=2mn+n^{2}$$
Your examples are the cases $(m,n)=(3,1)$ and $(5,1)$ respectively.
A: I'll start at
Will Jagy's hint.
If
$c^2
=a^2+ab+b^2
$,
$\begin{array}\\
c^4
&=a^4+a^2b^2+b^4+2a^3b+2a^2b^2+2ab^3\\
&=a^4+3a^2b^2+b^4+2a^3b+2ab^3\\
\text{so}\\
2c^4
&=2a^4+6a^2b^2+2b^4+4a^3b+4ab^3\\
&=a^4+b^2+a^4+6a^2b^2+b^4+4a^3b+4ab^3\\
&=a^4+b^4+a^4+4a^3b+6a^2b^2+4ab^3+b^4\\
&=a^4+b^4+(a+b)^4\\
\end{array}
$
Yep.
A: You have a disguised version of triangles with integer sides and one $120^\circ$ angle. These are
$$ 3,5,7 $$
$$ 7,8,13 $$
$$ 5,16,19$$
$$ 11,24, 31,  $$
which solve
$$ a^2 + ab + b^2 = c^2. $$
Square both sides and then double both sides and you get your identities.
These can be generated by a coprime pair of number $m,n$ with
$$ a = m^2 - n^2 $$
$$ b = 2mn+n^2  $$
$$ c = m^2 + mn + n^2 $$
