Consider the indefinite diagonal ternary form $$q(x,y,z)= 2 x^2 + 5 y^2 - 10 z^2$$ Based on numerical experience, I found that any given number of the form 5t+2 is represented either by $q$ or $-q$.
In other others, for any given $t$ there exist $x,y,z$ such that $q(x,y,z) = 5t+2$ or $q(x,y,z) = -5t-2$. Does anyone know a general method for proving this?
There seems to be a big literature on the genus of definite forms, but I didn't find to many references for indefinite ones.
NOTE: the method below will work for any $a x^2 + b y^2 - ab z^2$ with $\gcd(a,b)=1$ and both squarefree. It is a theorem using Hilbert's Norm Residue symbol that the number of primes for which an indefinite ternary form is anisotropic is even, therefore none, or two primes as below, or four, and so on. Oh, even when $a,b$ are both odd, the prime $2$ needs to be checked, and is part of the Hilbert thing.
i will check some things, but let me say the nature of the answer quickly, edit later. Your form is alone in its genus. This follows from a 1975 article by Hsia in English, as well as a 1956 article by Kneser. The point is that, for your "diagonal" form, the product of the coefficients is not divisible by $5^3$ or any $p^3$ for odd prime $p,$ and it is not divisible by $16; $ your product is not divisible by $8.$ I will need to check the exact power of $2$ for indefinite ternaries; the famous 1951 counterexample of Siegel is $x^2 - 2 y^2 + 64 z^2,$ but I suspect that does not show the sharp power of $2.$ For positive forms, the genus of $x^2 + y^2 + 16 z^2$ splits into two spinor genera, this goes back to Jones and Pall 1939. Both are exercises in CASSELS, $x^2 + y^2 + 16 z^2$ on page 252, $x^2 - 2 y^2 + 64 z^2,$ on page 253.
I have left out a good deal of detail; the simple result can be proved entirely formally. Finding the congruence obstructions is not difficult, I do like to confirm by computer. Confirming a form is anisotropic in the $p$-adic numbers for some $p$ is also not so bad. The big theorem is that every number not excluded by local considerations really is represented. This follows form being alone in its genus, general result by Jones, quantitative by Siegel.
References: two books by Leonard Eugene Dickson discuss indefinite forms, Modern Elementary theory of Numbers (1939), Studies in the Theory of Numbers (1930). All you will need for methods. Survey article by Schulze-Pillot (2004) Representation by integral quadratic forms- a survey in Contemporary Mathematics number 344, Algebraic and Arithmetic Theory of Quadratic Forms pages 303-321. He refers, in turn, to Hsia, Spinor Norms of local integral rotations, I, Pacific Journal of Mathematics volume 57, number 1, pages 199-206. It was actually Kneser 1956. Alright, it seems there are sufficient conditions for a genus of indefinite ternaries to have just one class in Dickson Studies (1930), sections 33-34, pages 54-60. Probably not all that intricate for diagonal ternaries.
This means that you just need to find the numbers that are permitted by "congruence obstacles." The traditional phrase for these prohibited numbers was "the progressions." All numbers not so prohibited really will be integrally represented by your $2 x^2 + 5 y^2 - 10 z^2.$
First, it is anisotropic in both the $5$-adic numbers and the $2$-adic. It is equivalent $\pmod {16}$ to the positive form $2 u^2 + 5 v^2 + 6 w^2,$ and is not allowed to be equal to $4^k (8n+1).$ To fill in the gaps here, we must find out a power of $2$ such that $2 u^2 + 5 v^2 - 10 w^2 \equiv 0 \pmod {2^e}$ implies that $u,v,w$ are all even. I think $e=4$ works.
The prime $5$ is easier. If $2 u^2 + 5 v^2 - 10 w^2 \equiv 0 \pmod {25},$ then $u,v,w$ are all divisible by $5.$ Then $2 u^2 + 5 v^2 - 10 w^2 \neq 25^k (5n \pm 1).$
That should do it. The missing numbers are either $4^k (8n+1)$ for this form, or $4^k (8n+7)$ for its negative.
YEP. All works. I had it write out all numbers (but not $\pm 1 \pmod 5$) not represented between $-1000$ and $1000.$ For each number $n$ not represented, I divided out by $4$ until the result was no longer divisible by $4,$ then printed the remainder $\pmod 8.$ I also divided out by $25$ until the result was no longer divisible by $25,$ then printed the remainder $\pmod 5.$ The $8$ remainder was always $1.$ Always $1.$
=========================================