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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.

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    $\begingroup$ Every integer not twice an odd number is $c^2-x^2$, so every integer is $c^2-x^2$ or $1^2+c^2-x^2$. $\endgroup$ Jul 31, 2013 at 23:29

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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

$$ \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & -1 & 1 \\ 1 & -1 & 1 \end{array} \right) \left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & -1 & -1 \\ 1 & 1 & 1 \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right). $$

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  • $\begingroup$ What's the integer-invertible linear change of variables which gives my original $b^2+c^2-x^2$? I'll happily mark you as the answer since you have a neat extension (but I'm scared to look at that paper and just trusting you). Also, per my original motivation, do you have any non-universal quaternary $M(a^2+b^2+c^2)-Nx^2$ forms where M and N are coprime integers? I know there's a good question in here somewhere... $\endgroup$
    – bobuhito
    Aug 1, 2013 at 23:47
  • $\begingroup$ @bobuhito, $3 (w^2 + x^2 + y^2) - z^2 \neq 1 \pmod 3.$ $\endgroup$
    – Will Jagy
    Aug 1, 2013 at 23:51
  • $\begingroup$ @bobuhito or $ w^2 + x^2 + y^2 - 8 z^2 \neq 7 \pmod 8. $ As far as good questions, you are looking for $M(a^2 + b^2 + c^2) - N x^2$ that is in a genus of at least two spinor genera and this is an irregular one. I'm not sure that is possible. So I suspect you are stuck with obvious $p$-adic obstructions. If I find an example of the type you ask about that is nontrivial i will let you know. $\endgroup$
    – Will Jagy
    Aug 2, 2013 at 0:02
  • $\begingroup$ Thanks, you are brilliant and fast and use terms beyond me. By the way, I was originally looking for a quadratic form that ranges all but a finite set of integers. The closer the quadratic form looks to the Minkowski metric, the better...then this impossible set of integers is kind of like the electron and other excitations of spacetime. Do you have a candidate quadratic form, or can you already tell me that's impossible? $\endgroup$
    – bobuhito
    Aug 2, 2013 at 0:21
  • $\begingroup$ @bobuhito, impossible, as indefinite forms in at least three variables are spinor regular, so either they are universal or they have a $p$-adic obstruction or a spinor obstruction, infinite in either case. Quite easy for positive forms to miss a finite set of numbers. $\endgroup$
    – Will Jagy
    Aug 2, 2013 at 0:25
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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).

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    $\begingroup$ I often boast about how humble I am. $\endgroup$ Aug 2, 2013 at 0:39

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