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: 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).
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.
A: Note that $x^2+y^2-z^2=x^2+(y+z)(y-z)$. The $(y+z)(y-z)$ term is quite versatile. It represents lots (but not quite all) of integers.
A: 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 $x,y,z\neq 0$.
