Smallest Square which is Sum of 3 Fourth Powers

Theorem
The smallest positive integer whose square is the sum of $3$ fourth powers is $481$:
 * $481^2 = 12^4 + 15^4 + 20^4$

Proof
The smallest solution must be fourth powers of coprime integers, otherwise dividing by their GCD would yield a smaller solution.

By Fermat's Right Triangle Theorem, $x^4 + y^4 = z^2$ has no solutions.

Thus none of the fourth powers is $0$.

Consider the equation $x^4 + y^4 + z^4 = n^2$.

By Square Modulo 4, since fourth powers are squares, at most one of $x, y, z$ is odd.

If none of them are odd, the numbers are not coprime.

Therefore there is exactly one odd number.

By Fermat's Little Theorem, for $3 \nmid a$:
 * $a^2 \equiv 1 \pmod 3$

Therefore by Congruence of Powers:
 * $a^4 \equiv 1^2 \equiv 1 \pmod 3$

For $3 \divides a$:
 * $a^2 \equiv a^4 \equiv 0 \pmod 3$

Therefore we cannot have two of $x, y, z$ not divisible by $3$.

By our coprimality condition we cannot have $3 \divides x, y, z$.

Hence either $3 \nmid x, y, z$ or exactly one of $x, y, z$ is not divisible by $3$.

Similarly, by Fermat's Little Theorem, for $5 \nmid a$:
 * $a^4 \equiv 1 \pmod 5$

For $5 \divides a$:
 * $a^4 \equiv 0 \pmod 5$

By Square Modulo 5:
 * $a^2 \equiv 0, 1, 4 \pmod 5$

Therefore there is exactly one number not divisible by $5$.

Since $\sqrt {481} < 22$, we only need to check fourth powers of numbers less than $22$, satisfying the criteria above.

Since there are exactly two multiples of $2$ and $5$ among $x, y, z$, at least one of them must be a multiple of $10$.

Suppose $5 \nmid x$, $5 \divides y$ and $10 \divides z$.

Case $1$: $z = 10$
The case $y = 20$ will be equivalent to Case $2$.

Suppose $y = 15$.

By the criteria above, $x$ is divisible by $2$ and $3$.

Also $x < 22$, so we only need to check $x = 6, 12, 18$.

Suppose $y = 10$.

By the criteria above, $x$ is not divisible by $2$ or $3$.

Also $x < 22$, so we only need to check $x = 1, 7, 11, 13, 17, 19$.

Suppose $y = 5$.

By the criteria above, $x$ is divisible by $2$ but not $3$.

Also $x < 22$, so we only need to check $x = 2, 4, 8, 14, 16$.

Case $2$: $z = 20$
Suppose $y = 20$. Then $x^4 + y^4 + z^4 > 20^4 + 20^4 = 320 \, 000 > 481^2$.

Suppose $y = 15$.

By the criteria above, $x$ is divisible by $2$ and $3$.

Also $x < \sqrt [4] {481^2 - 20^4 - 15^4} = 12$, so we only need to check $x = 6$.

Suppose $y = 10$.

By the criteria above, $x$ is not divisible by $2$ or $3$.

Also $x < \sqrt [4] {481^2 - 20^4} < 17$, so we only need to check $x = 1, 7, 11, 13$.

Suppose $y = 5$.

By the criteria above, $x$ is divisible by $2$ but not $3$.

Also $x < \sqrt [4] {481^2 - 20^4} < 17$, so we only need to check $x = 2, 4, 8, 14, 16$.

And we see that none of the above cases yield any smaller solutions.

Hence the result.