Divisibility of Sum of 3 Fourth Powers

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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:

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


From Fourth Power Modulo 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.


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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 $R$


This article, or a section of it, needs explaining.
In particular: The above statement needs to be gone into in more detail, probably on a separate page.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.
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$\blacksquare$


Historical Note

This result was established by Leonhard Paul Euler.


Sources

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