I've come across several number theory problems concerning solutions to the following kinds of equations involving conic sections/quadratic forms. In three variables for instance.
$$30x^2 +5y^2 +17z^2 -15xy -70xz +50yz = 0$$
I've found several classification theorems and tricks to determine whether or not there exist integer solutions, this mainly consists either in brute-force methods, trying to split the former into systems of equations, or employing other theorems.
There doesn't seem to be any general method for trying to solve equations of this type, but I would like to know of strategies that can be employed in solving these equations once one has determined that there exist solutions.
Any help would be appreciated
No integer solutions. If there are any solutions with $x,y,z$ not all zero, divide out by the gcd of them, the result is $30x^2 +5y^2 +17z^2 -15xy -70xz +50yz = 0$ with the new restriction that $\gcd(x,y,z) = 1,$ still $x,y,z$ not all zero.
ASSUMING not all zero and gcd one...
IF $$30x^2 +5y^2 +17z^2 -15xy -70xz +50yz \equiv 0 \pmod {169}$$ then $x,y,z$ are all divisible by $13.$ This is a contradiction of the assumption. So, only the trivial solution.
Same idea for prime $5$ and $$30x^2 +5y^2 +17z^2 -15xy -70xz +50yz \equiv 0 \pmod {3125}$$
One theorem, for an indefinite integral quadratic form, dimension 3, the number of (finite) primes at which the form is anisotropic is even; this is the product relation for the Hilbert Norm Residue symbol.
I'll try to find a more transparent method. The theorem to figure out is Legendre's about (diagonal) ternary quadratic forms. I like the treatment in Cassels. Statement on page 80. However, it is not until page 82 that he gives a Note to the effect that one may ignore the prime 2 if the form is indefinite. This follows from the global relation.
Wednesday morning:
$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 4 } & 1 & 0 \\ - \frac{ 7 }{ 6 } & \frac{ 26 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 60 & 0 & 0 \\ 0 & \frac{ 25 }{ 4 } & 0 \\ 0 & 0 & - \frac{ 650 }{ 3 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 4 } & - \frac{ 7 }{ 6 } \\ 0 & 1 & \frac{ 26 }{ 5 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 60 & - 15 & - 70 \\ - 15 & 10 & 50 \\ - 70 & 50 & 34 \\ \end{array} \right) $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ f =30x^2 +5y^2 +17z^2 -15xy -70xz +50yz $$
$$ f = 30 \left( x - \frac{y}{4} -\frac{7z}{6} \right)^2 + \frac{25}{8} \left( y + \frac{26z}{5} \right)^2 - \frac{325}{3} z^2 $$
If $f = 0$ in integer $x,y,z$ we have produced a rational null vector $(u,v,w) $ for $30 u^2 + \frac{25}{8} v^2 - \frac{325}{3} w^2 . \; \;$ Divide all the coefficients by $5$ and we have a null vector for $6 u^2 + \frac{5}{8} v^2 - \frac{65}{3} w^2 . \; \;$ Let $v = 4 v_1$ so that $6 u^2 + 10 v_1^2 - \frac{65}{3} w^2 =0 \; \; .$ Let $w = 3 w_1$ for $6 u^2 + 10 v_1^2 - 195 w_1^2 =0 \; \; .$ Let $u = 5 u_1$ for $150 u_1^2 + 10 v_1^2 - 195 w_1^2 =0 \; \; .$ Divide coefficients by $5$ again for $30 u_1^2 + 2 v_1^2 - 39 w_1^2 =0 \; \; .$ Let$v_1 = 3 v_2$ for $30 u_1^2 + 18 v_2^2 - 39 w_1^2 =0 \; \; .$ Divide coefficients by $3$ for $10 u_1^2 + 6 v_2^2 - 13 w_1^2 =0 \; \; .$ Let $w_1 = 2 w_2 $ for $10 u_1^2 + 6 v_2^2 - 52 w_2^2 =0 \; \; .$ Divide coefficients by $2$ for
$$5 u_1^2 + 3 v_2^2 - 26 w_2^2 =0 \; \; .$$
This is finally in correct shape for Legendre's Theorem, pages 80-82 in Cassels, Rational Quadratic Forms. Page 42 in Serre.
However, I will just show the calculation: if there is a rational null vector $(u_1,v_2, w_2)$ with at least one of these nonzero, we may multiply the three rational multiple by their least common denominator, the result is now all integers, call them $(p,q,r)$ not all zero with $$5 p^2 + 3 q^2 - 26 r^2 =0 \; \; .$$
Furthermore, we may divide out by any common factor to arrange (not all zero and) $ \gcd(p,q,r) = 1.$ We do a calculation to show that all three must be divisible by $5.$ $$5 p^2 + 3 q^2 - 26 r^2 \equiv 3 q^2 - 26 r^2 \equiv 3 q^2 +4 r^2 \pmod 5$$
The squares are $t^2 \equiv 0,1,4 \pmod 5,$ with the $0$ occurring only when $ t \equiv 0 \pmod 5$
The values of $3q^2 \equiv 0,3,2 \pmod 5,$ with the $0$ occurring only when $ q \equiv 0 \pmod 5$ The values of $4r^2 \equiv 0,4,1 \pmod 5,$ with the $0$ occurring only when $ r \equiv 0 \pmod 5$ The two numbers add up to a multiple of $5$ only when both are zero; that is, $$ 3 q^2 - 26 r^2 \equiv 0 \pmod 5 $$ only when both $q,r$ are divisible by $5,$ so that $q = 5 q_1$ and $r = 5 r_1$ Then we have $$5 p^2 + 3 q^2 - 26 r^2 = 5 p^2 + 75 q_1^2 - 650 r_1^2 = 0$$ and $$ p^2 + 15 q_1^2 - 130 r_1^2 = 0$$ It follows immediately that $p \equiv 0 \pmod 5.$ However, we have reached all of $p,q,r$ divisible by $5,$ which contradicts the assumption the $\gcd(p,q,r) = 1$ And that contradicts the assumption that we have integers , not all zero, with $$ 30x^2 +5y^2 +17z^2 -15xy -70xz +50yz = 0. $$