# Divisibility of Sum of 3 Fourth Powers

## Theorem

Let $n \in \Z_{\ge 0}$ be the sum of three $4$th powers.

Then:

$n$ is divisible by $5$ if and only if all three addends are also divisible by $5$
$n$ is divisible by $29$ if and only if all three addends are also divisible by $29$.

## Proof

Let $n = a^4 + b^4 + c^4$ for $a, b, c \in \Z$.

### Necessary Condition

Let $a$, $b$ and $c$ all be divisible by either $5$ or $29$.

That is:

$a, b, c \equiv 0 \pmod 5$

or:

$a, b, c \equiv 0 \pmod {29}$

We have that for integer $p$:

 $\ds a, b, c$ $\equiv$ $\ds 0$ $\ds \pmod p$ $\ds \leadsto \ \$ $\ds a^4, b^4, c^4$ $\equiv$ $\ds 0$ $\ds \pmod p$ Congruence of Powers $\ds$ $\equiv$ $\ds 0$ $\ds \pmod p$ Modulo Addition is Well-Defined $\ds \leadsto \ \$ $\ds p$ $\divides$ $\ds a^4 + b^4 + c^4$ where $\divides$ denotes divisibility

Hence if all three addends are divisible by either $5$ or $29$, then so is the sum.

$\Box$

### Sufficient Condition

Let $n$ be divisible by $5$.

Then:

$a^4 + b^4 + c^4 \equiv 0 \mod 5$
$x \in \Z_5 \implies x^4 \in \set {0, 1}$

So:

$a^4 + b^4 + c^4 \equiv 0 \mod 5 \implies a, b, c \equiv 0 \mod 5$

$\Box$

Let $n$ be divisible by $29$.

Then $a^4 + b^4 + c^4 \equiv 0 \mod {29}$.

But for $x \in \Z_{29}$, $x^4 \in \set {0, 1, 7, 16, 20, 23, 24, 25} = R$ necessarily.

So:

$a^4 + b^4 + c^4 \equiv 0 \mod {29} \implies a, b, c \equiv 0 \mod {29}$

as $29$ and its non-zero multiples are not partitioned by any 3 elements in $R$

$\blacksquare$

## Historical Note

This result was established by Leonhard Paul Euler.