Four-Square Theorem for Negative Coefficient What integers are not in the range of $a^2+b^2+c^2-x^2$ (for all integer combinations of a, b, c, and x)?  This form is similar to that of Lagrange's Four-Square Theorem, for which the answer would be "none".  The generalization by Ramanujan only seems to cover non-negative coefficients.
 A: Since Gerry Myerson is too humble to take answer credit, I'll
make it even simpler and close out my stupid question:
$0^2+(x+1)^2-x^2=2x+1$ ranges all odds, so $1^2+(x+1)^2-x^2$ ranges all evens.
Just 3 integers are enough (a is 0, b is 0 or 1, c is x+1).
A: With indefinite forms, it is possible for ternary forms to be universal. Indeed, all are known. References are given in Modern Elementary Theory of Numbers by Leonard Eugene Dickson, (1939).  With any  integer $M$ and any odd $N,$ they are equivalent to (by an invertible linear change of variables) one of four:
$$  xy - M z^2, $$
$$  2xy - N z^2, $$
$$  2  xy + y^2 - N z^2, $$
$$  2  xy + y^2 - 2 N z^2. $$
If you allow $M=0$ you get a universal binary.
There are universal quaternary forms without such a universal ternary "section," for example $$  w^2 - 2 x^2 + 3 y^2 - 6 z^2.  $$ 
EEEEEDDDDIITTTTT:
$2xy+ z^2$ is integrally equivalent to $a^2 + b^2 - c^2,$ by
$$  
 \left(  \begin{array}{ccc}
  1  &  0 & 1  \\
   0   &  -1 & 1 \\
   1 & -1 & 1  
\end{array} 
  \right) 
 \left(  \begin{array}{ccc}
  0  &  1 & 0  \\
   1   &  0 & 0 \\
   0 & 0 & 1  
\end{array} 
  \right) 
 \left(  \begin{array}{ccc}
  1  &  0 & 1  \\
   0   &  -1 & -1 \\
   1 & 1 & 1  
\end{array} 
  \right)  
 =
 \left(  \begin{array}{ccc}
  1  &  0 & 0  \\
   0   &  1 & 0 \\
   0 & 0 & -1  
\end{array} 
  \right).  
 $$
