every integer can be represented in the form $x^2+y^2-z^2$ Every integer can be represented in the form $x^2+y^2-z^2$ and show that $6$ actually requires all three terms.
I put 


*

*$z=y+1$ 

*$x=n^2+3$ 

*$y=3n^2+4+(n^4-n)/2$


what does it mean that $6$ actually requires all three terms?

Notice all odd numbers are:
$1 = 1^2 - 0^2$
$3 = 2^2 - 1^2$
$5 = 3^2 - 2^2$ etc..
so if odd numbers are $\{o_1,o_2,...\}$ then $o_n=n^2-(n-1)^2$.
Now even numbers $\{e_1,e_2,...\}$ are simply $e_n=o_n+1^2$.

regarding the number 6, if any shorter representation exists, it must be one of the following:
1) $x^2 + y^2$ - you can easily see no such exists.
2) $x^2 - z^2$ - if this is it, $x,y\in\{1,2,3,4\}$ otherwise their difference will be greater than 6, but again, no such combination exists...
Hence all 3 must be non-zero
