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?

It means that if $x,y,z$ are such that
$$x^2+y^2-z^2=6$$
then $x,y,z$ are all non-zero.
You can prove this by looking at the equation mod $4$.
You know that a square is always $0$ or $1$ mod $4$, and $6\equiv 2\pmod 4$.
So if $x^2+y^2-z^2=6$, then $z$ must be $0$ mod $4$, otherwise you are not able to get to $2$ mod $4$.
But $6$ can't be a sum of two squares mod $4$ because $3\mid 6$ (look there).
