An identity which applies to all of the natural numbers

Prove that any natural number n can be written as $$n=a^2+b^2-c^2$$ where $a,b,c$ are also natural.

Solution 1:

Consider $n\ge 6$.

If $n$ is odd, $n=2m+1$, then $$ n = 2m+1 = 2^2 + (m-1)^2 - (m-2)^2; $$ If $n$ is even, $n=2m$, then $$ n = 2m = 1^2 + m^2 - (m-1)^2; $$

Small $n$:
$1=1^2+1^2-1^2$,
$2=3^2+3^2-4^2$,
$3=4^2+6^2-7^2$,
$4=2^2+1^2-1^2$,
$5=4^2+5^2-6^2$.

Solution 2:

If you consider the case $b=c+1$, you get $b^2-c^2=2c+1$, which can be any odd natural number $\ge3$. Thus with $a=1$ you reach all even $n\ge4$ and with $a=2$ all odd $n\ge 7$. This leaves only the cases $n\in\{1,2,3,5\}$ open. Can you find solutions for these $n$? (Hint: Try $b=c-1$).