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?

Any integer number that is not of the form $4m+2$ can be represented as a difference of two squares, since $n=y^2-z^2=(y-z)(y+z)$ has a solution as soon as $n$ has a divisor $d$ such that $d$ and its complementary divisor $\frac{n}{d}$ have the same parity. That is the same as stating "$n$ is not twice an odd number". It follows that $n=x^2+y^2-z^2$ always has a solution with $x=0$ or $x=1$.
$6$ is a number of the form $4m+2$, hence $x\neq 0$, and it is not the sum of two squares, since $3\mid 6$ but $3^2\nmid 6$. It follows that $6=x^2+y^2-z^2$ has a lot of solutions (like $6=5^2+9^2-10^2$) but for all of them $x,y,z\neq 0$.
