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 $5$ all three addends are also divisible by $5$
 * $n$ is divisible by $29$ 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 $p$:

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

Sufficient Condition
Let $n$ be divisible by $5$.

Then $a^4 + b^4 + c^4 \equiv 0 \mod 5$.

But for $x \in \Z_5$, $x^4 \in \set {0, 1}$ necessarily.

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

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 $R$