Q:

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?

A:

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

A:

Every odd number is of the form $2n+1=(n+1)^2-n^2=(n+1)^2-n^2+0^2$ and every even integer is of the form $2n=(n+1)^2-n^2-1^2$.
Now the second part:
Let, $6=x^2+y^2$ i.e $z=0$.. Now, $x,y\ne1$ as if one of them is one the other one will be $\sqrt{5}$. $x,y\ne2$ as if one of them is $2$ other one will be $\sqrt{2}$. We don't need to check it for $3$ as the squares of integers is always positive.
Now, assume that $6=x^2-z^2$ i.e $y=0$ then $6=(x+z)(x-z)$ . Since $6=-1*-6=6*1=2*3$ and sum of any these two pairwise factors is always odd. While sum of $x+z,x-z$ is even($2x$). A contradiction.

