Sums of Fourth Powers

Question

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?

Answer 1

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 $$ a = m^2 - n^2 $$ $$ b = 2mn+n^2 $$ $$ c = m^2 + mn + n^2 $$

Answer 2

First, observe $$\begin{align*}a^{4}+b^{4}+(a+b)^{4}&=2a^{4}+4a^{3}b+6a^{2}b^{2}+4ab^{3}+2b^{4}\\&=2(a^{4}+2a^{3}b+3a^{2}b^{2}+2ab^{3}+b^{4})\\&=2(a^{2}+ab+b^{2})^{2}\end{align*}$$ 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 $m,n$. $$\begin{align*}a&=m^{2}-n^{2}\\b&=2mn+n^{2}\end{align*}$$ Your examples are the cases $(m,n)=(3,1)$ and $(5,1)$ respectively.

Answer 3

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.