Let $n > 169$ be an integer. Show that $n$ can be written as the sum of five positive squares.
Hint: Consider $n = m + 169$.
Note, a positive square is greater than 0, that is, not 0^2.
By Lagrange’s Sum of Four Squares Theorem (8.14), we know that every positive integer can be expressed as a sum of four squares (some of which can be zero).
And so we can write a positive integer $m$ as
$$ m = a^2 + b^2 + c^2 + d^2 $$
We define $n=m+169$, and noting that $169=13^2$
$$ n = a^2 + b^2 + c^2 + d^2 + 13^2 $$
We now consider the cases where some of $a,b,c,d$ are 0:
- One of $a,b,c,d$ is 0. Without loss of generality, $d=0$.
- Two of $a,b,c,d$ are 0. Without loss of generality, $c=d=0$.
- Three of $a,b,c,d$ are 0. Without loss of generality, $b=c=d=0$.
All four $a,b,c,d$ can't be zero because this would give $n=169$ which contradicts $n>169$. Let's consider each case in turn.
Case One
We have, noting that $13^2 = 12^2 + 5^2$
$$ n = a^2 + b^2 + c^2 + 0^2 + (13^2) = a^2 + b^2 + c^2 + 12^2 + 5^2 $$
In this case $n$ can be expressed as the sum of five positive squares.
Case Two
We have, noting that $13^2 = 3^2 + 4^2 + 12^2$
$$ n = a^2 + b^2 + 0^2 + 0^2 + (13^2) = a^2 + b^2 +3^2 + 4^2 + 12^2 $$
In this case $n$ can be expressed as the sum of five positive squares.
Case Three
We have, noting that $13^2 = 1^2 + 2^2 + 8^2 +10^2$
$$ n = a^2 + 0^2 + 0^2 + 0^2 + (13^2) = a^2 + 1^2 + 2^2 + 8^2 +10^2 $$
In this case $n$ can be expressed as the sum of five positive squares.
In all cases we have shown that $n$ can be expressed as the sum of five positive squares.