Integer as Sum of 5 Non-Zero Squares

Theorem
Let $n \in \Z$ be an integer such that $n > 33$.

Then $n$ can be expressed as the sum of $5$ non-zero squares.

Proof
From Lagrange's Four Square Theorem, every positive integer can be expressed as the sum of $4$ squares, some of which may be zero.

The existence of positive integers which cannot be expressed as the sum of $4$ non-zero squares is noted by the trivial examples $1$, $2$ and $3$.

Thus Lagrange's Four Square Theorem can be expressed in the form:
 * $(1): \quad$ Every positive integer can be expressed as the sum of $1$, $2$, $3$ or $4$ non-zero squares.

We note the following from 169 as Sum of up to 155 Squares:

Let $n > 169$.

Then $n$ can be expressed as:
 * $n = m + 169$

where $m \ge 1$.

From $(1)$, $m$ can be expressed as the sum of sum of $1$, $2$, $3$ or $4$ non-zero squares.

Thus at least one of the following holds:
 * $m = a^2$
 * $m = a^2 + b^2$
 * $m = a^2 + b^2 + c^2$
 * $m = a^2 + b^2 + c^2 + d^2$

Thus one of the following holds:

It remains to be shown that of the positive integers less than $169$, all but the following can be expressed in this way:
 * $1, 2, 3, 4, 6, 7, 9, 10, 12, 15, 18, 33$

Note that by Integer as Sum of Three Squares, all integers not of the form:
 * $4^n \paren {8 m + 7}$

can be written as a sum of $1$, $2$ or $3$ non-zero squares.

Also note that:

Similar to the above, for $x = y + 18$ where $y \ne 4^n \paren {8 m + 7}$, at least one of the following holds:

the ineligible $0 < y < 151$ are:
 * $7, 15, 23, 28, 31, 39, 47, 55, 60, 63, 71, 79, 87, 92, 95, 103, 111, 112, 119, 124, 127, 135, 143$

with corresponding $x$:
 * $25, 33, 41, 46, 49, 57, 65, 73, 78, 81, 89, 97, 105, 110, 113, 121, 129, 130, 137, 142, 145, 153, 161$

for $x > 18$.

Similarly, for $45$:

So if we can write $x = y + 45$ where $y \ne 4^n \paren {8 m + 7}$, $x$ can be expressed as a sum of $5$ non-zero squares.

The ineligible $x > 45$ for $0 < y < 124$ are:
 * $52, 60, 68, 73, 76, 84, 92, 100, 105, 108, 116, 124, 132, 137, 140, 148, 156, 157, 164$

Comparing both lists, we only need to check:
 * $x < 18$ and $x = 25, 33, 41, 73, 105, 137$

And we have:

while for the rest:
 * $1, 2, 3, 4 < 5 \times 1^2$
 * $5 \times 1^2 < 6, 7 < 2^2 + 4 \times 1^2$
 * $2^2 + 4 \times 1^2 < 9, 10 < 2 \times 2^2 + 3 \times 1^2$

$12, 15, 18, 33$ are divisible by $3$.

By Square Modulo 3, $n^2 \equiv 0$ or $1 \pmod 3$.

We must require the $5$ non-zero squares to be equivalent to:
 * $0, 0, 1, 1, 1 \pmod 3$

or
 * $0, 0, 0, 0, 0 \pmod 3$

The smallest non-zero square divisible by $3$ is $3^2 = 9$.

The sum of the squares must therefore be greater than:
 * $3^2 + 3^2 = 18$

hence $12, 15, 18$ cannot be expressed as the sum of $5$ non-zero squares.

Since $6^2 > 33$, we must have $33 = 3^2 + 3^2 + a^2 + b^2 + c^2$.

But $33 - 3^2 - 3^2 = 15$ cannot be expressed as the sum of $3$ non-zero squares, as it is of the form $4^n \paren {8 m + 7}$.

This proves the theorem.