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?

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.
