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:
\(\ds 169\) | \(=\) | \(\ds 13^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 12^2 + 5^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 12^2 + 4^2 + 3^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 8^2 + 8^2 + 5^2 + 4^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 8^2 + 8^2 + 4^2 + 4^2 + 3^2\) |
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:
\(\ds n\) | \(=\) | \(\ds a^2 + b^2 + c^2 + d^2 + 13^2\) | ||||||||||||
\(\ds n\) | \(=\) | \(\ds a^2 + b^2 + c^2 + 12^2 + 5^2\) | ||||||||||||
\(\ds n\) | \(=\) | \(\ds a^2 + b^2 + 12^2 + 4^2 + 3^2\) | ||||||||||||
\(\ds n\) | \(=\) | \(\ds a^2 + 8^2 + 8^2 + 5^2 + 4^2\) |
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:
\(\ds 18\) | \(=\) | \(\ds 3^2 + 3^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4^2 + 1^2 + 1^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3^2 + 2^2 + 2^2 + 1^2\) |
Similar to the above, for $x = y + 18$ where $y \ne 4^n \paren {8 m + 7}$, at least one of the following holds:
\(\ds n\) | \(=\) | \(\ds a^2 + b^2 + c^2 + 3^2 + 3^2\) | ||||||||||||
\(\ds n\) | \(=\) | \(\ds a^2 + b^2 + 4^2 + 1^2 + 1^2\) | ||||||||||||
\(\ds n\) | \(=\) | \(\ds a^2 + 3^2 + 2^2 + 2^2 + 1^2\) |
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 $45$:
\(\ds 45\) | \(=\) | \(\ds 6^2 + 3^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 5^2 + 4^2 + 2^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4^2 + 4^2 + 3^2 + 2^2\) |
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:
\(\ds 5\) | \(=\) | \(\ds 1^2 + 1^2 + 1^2 + 1^2 + 1^2\) | ||||||||||||
\(\ds 8\) | \(=\) | \(\ds 2^2 + 1^2 + 1^2 + 1^2 + 1^2\) | ||||||||||||
\(\ds 11\) | \(=\) | \(\ds 2^2 + 2^2 + 1^2 + 1^2 + 1^2\) | ||||||||||||
\(\ds 13\) | \(=\) | \(\ds 3^2 + 1^2 + 1^2 + 1^2 + 1^2\) | ||||||||||||
\(\ds 14\) | \(=\) | \(\ds 2^2 + 2^2 + 2^2 + 1^2 + 1^2\) | ||||||||||||
\(\ds 16\) | \(=\) | \(\ds 3^2 + 2^2 + 1^2 + 1^2 + 1^2\) | ||||||||||||
\(\ds 17\) | \(=\) | \(\ds 2^2 + 2^2 + 2^2 + 2^2 + 1^2\) | ||||||||||||
\(\ds 25\) | \(=\) | \(\ds 3^2 + 2^2 + 2^2 + 2^2 + 2^2\) | ||||||||||||
\(\ds 41\) | \(=\) | \(\ds 4^2 + 4^2 + 2^2 + 2^2 + 1^2\) | ||||||||||||
\(\ds 73\) | \(=\) | \(\ds 6^2 + 5^2 + 2^2 + 2^2 + 2^2\) | ||||||||||||
\(\ds 105\) | \(=\) | \(\ds 6^2 + 6^2 + 5^2 + 2^2 + 2^2\) | ||||||||||||
\(\ds 137\) | \(=\) | \(\ds 10^2 + 5^2 + 2^2 + 2^2 + 2^2\) |
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.
$\blacksquare$
Sources
- Feb. 1993: Kelly Jackson, Francis Masat and Robert Mitchell: Extensions of a Sums-of-Squares Problem (Math. Mag. Vol. 66, no. 1: pp. 41 – 43) www.jstor.org/stable/2690474
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $33$