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: {{begin-eqn}} {{eqn | l = 169       | r = 13^2       | c =  }} {{eqn | r = 12^2 + 5^2       | c =  }} {{eqn | r = 12^2 + 4^2 + 3^2       | c =  }} {{eqn | r = 8^2 + 8^2 + 5^2 + 4^2       | c =  }} {{eqn | r = 8^2 + 8^2 + 4^2 + 4^2 + 3^2       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = n       | r = a^2 + b^2 + c^2 + d^2 + 13^2       | c =  }} {{eqn | l = n       | r = a^2 + b^2 + c^2 + 12^2 + 5^2       | c =  }} {{eqn | l = n       | r = a^2 + b^2 + 12^2 + 4^2 + 3^2       | c =  }} {{eqn | l = n       | r = a^2 + 8^2 + 8^2 + 5^2 + 4^2       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = 18       | r = 3^2 + 3^2 }} {{eqn | r = 4^2 + 1^2 + 1^2 }} {{eqn | r = 3^2 + 2^2 + 2^2 + 1^2 }} {{end-eqn}} Similar to the above, for $x = y + 18$ where $y \ne 4^n \paren {8 m + 7}$, at least one of the following holds: {{begin-eqn}} {{eqn | l = n       | r = a^2 + b^2 + c^2 + 3^2 + 3^2       | c =  }} {{eqn | l = n       | r = a^2 + b^2 + 4^2 + 1^2 + 1^2       | c =  }} {{eqn | l = n       | r = a^2 + 3^2 + 2^2 + 2^2 + 1^2       | c =  }} {{end-eqn}} 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 $45$: {{begin-eqn}} {{eqn | l = 45       | r = 6^2 + 3^2 }} {{eqn | r = 5^2 + 4^2 + 2^2 }} {{eqn | r = 4^2 + 4^2 + 3^2 + 2^2 }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = 5       | r = 1^2 + 1^2 + 1^2 + 1^2 + 1^2       | c =  }} {{eqn | l = 8       | r = 2^2 + 1^2 + 1^2 + 1^2 + 1^2       | c =  }} {{eqn | l = 11       | r = 2^2 + 2^2 + 1^2 + 1^2 + 1^2       | c =  }} {{eqn | l = 13       | r = 3^2 + 1^2 + 1^2 + 1^2 + 1^2       | c =  }} {{eqn | l = 14       | r = 2^2 + 2^2 + 2^2 + 1^2 + 1^2       | c =  }} {{eqn | l = 16       | r = 3^2 + 2^2 + 1^2 + 1^2 + 1^2       | c =  }} {{eqn | l = 17       | r = 2^2 + 2^2 + 2^2 + 2^2 + 1^2       | c =  }} {{eqn | l = 25       | r = 3^2 + 2^2 + 2^2 + 2^2 + 2^2       | c =  }} {{eqn | l = 41       | r = 4^2 + 4^2 + 2^2 + 2^2 + 1^2       | c =  }} {{eqn | l = 73       | r = 6^2 + 5^2 + 2^2 + 2^2 + 2^2       | c =  }} {{eqn | l = 105       | r = 6^2 + 6^2 + 5^2 + 2^2 + 2^2       | c =  }} {{eqn | l = 137       | r = 10^2 + 5^2 + 2^2 + 2^2 + 2^2       | c =  }} {{end-eqn}} 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: :$0, 0, 1, 1, 1 \pmod 3$ 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 $15$ cannot be expressed as the sum of $3$ non-zero squares: :$15 < 4^2$ :$15 - 3^2 = 6$ is not the sum of $2$ squares :$15 > 3 \times 2^2$ The proves the theorem. {{qed}}