Let $F(X,Y,Z)=5X^2+3Y^2+8Z^2+6(YZ+ZX+XY)$. Find $(a,b,c) \in \mathbb{Z}^3$ not all divisible by $13$, such that $F(a,b,c)\equiv 0 \pmod{13^2}$.
Although it looks like a homework problem it is not. It is an exercise from Milne's Elliptic curves book and I am trying to read that book on my own. If someone could help me out then my conception would be little better. Thanks in advance.
On
First, a solution:
$x=3$, $y=49$, $z=1$ give: \begin{align} &5x^2+3y^2+8z^2+6(xy+yz+zx)\\ \mapsto&45+7203+1+6(147+49+3)\\ =&8450\\ =&50\cdot13^2 \end{align}
Now, the method:
It’s a conic section, and if you think $13$-adically, it either has no points at all over $\Bbb Q_{13}$ (or what’s the same thing here, over $\Bbb Z_{13}$) or it has infinitely many. So, as @tjf sagely suggested, if you can find a solution that’s good modulo $13$, you should be able to find one modulo $13^2$. I confess that I didn’t bother checking at this point that the thing is nonsingular, i.e. is neither a pair of intersecting lines nor a single line of multiplicity two.
I just went ahead and dehomogenized, setting $z=1$, to get $$ 5x^2+6xy+3y^2+6(x+y)=-8\,. $$ Since the discriminant of $5x^2+6xy+3y^2$ is $9-15\equiv7$ and $7$ is not a square modulo $13$, there are no points on the line at infinity. Drat, it’s an ellipse. I tried finding intersections with $x=0$, $y=0$, and $x=y$, but none of these three lines seems to intersect over $\Bbb F_{13}$. Drat again. But then I tried the line $x=-y$, and got a point $(3,-3)$ in the finite $\Bbb F_{13}$-plane, i.e. $x\equiv3$, $y\equiv10$, $z\equiv1$ as points of the homogeneous curve modulo $13$. From there it was an elementary Henselation to refine to a congruence modulo $169$.
$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 3 }{ 5 } & 1 & 0 \\ 0 & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 8 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 3 }{ 5 } & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & \frac{ 6 }{ 5 } & 0 \\ 0 & 0 & 5 \\ \end{array} \right) $$
OR $$\left( \begin{array}{rrr} 5 & 0 & 0 \\ - 3 & 5 & 0 \\ 0 & - 5 & 5 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 8 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & - 3 & 0 \\ 0 & 5 & - 5 \\ 0 & 0 & 5 \\ \end{array} \right) = \left( \begin{array}{rrr} 125 & 0 & 0 \\ 0 & 30 & 0 \\ 0 & 0 & 125 \\ \end{array} \right) $$
If we get $25 u^2 + 6 v^2 + 25 w^2 \equiv 0 \pmod {13^2}$ we have a linear transformation that takes that vector to one that is null for the original quadratic form. Since there is a Pythagorean triple (5,12,13) or $5^2 + 12 ^2 = 13 ^2,$ we use the matching coefficients 25 and take $u=5,v=0,w=12.$
$$ \left( \begin{array}{rrr} 5 & - 3 & 0 \\ 0 & 5 & - 5 \\ 0 & 0 & 5 \\ \end{array} \right) \left( \begin{array}{r} 5 \\ 0 \\ 12 \\ \end{array} \right) = \left( \begin{array}{r} 25 \\ -60 \\ 60 \\ \end{array} \right) $$ Divide the last vector by $5$ to get $$ \left( \begin{array}{r} 5 \\ -12 \\ 12 \\ \end{array} \right) $$ With $$ x = 5 , y = -12, z = 12, $$
$$ 5 x^2 + 3 y^2 + 8 z^2 + 6 (yz + z x + x y) = 845 = 5 \cdot 13^2 $$
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$$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$
$$ H = \left( \begin{array}{rrr} 5 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 8 \\ \end{array} \right) $$
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$$ E_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 3 }{ 5 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 3 }{ 5 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & \frac{ 3 }{ 5 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 5 & 0 & 3 \\ 0 & \frac{ 6 }{ 5 } & \frac{ 6 }{ 5 } \\ 3 & \frac{ 6 }{ 5 } & 8 \\ \end{array} \right) $$
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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & - \frac{ 3 }{ 5 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 3 }{ 5 } & - \frac{ 3 }{ 5 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & \frac{ 3 }{ 5 } & \frac{ 3 }{ 5 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & \frac{ 6 }{ 5 } & \frac{ 6 }{ 5 } \\ 0 & \frac{ 6 }{ 5 } & \frac{ 31 }{ 5 } \\ \end{array} \right) $$
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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 3 }{ 5 } & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & \frac{ 3 }{ 5 } & \frac{ 3 }{ 5 } \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & \frac{ 6 }{ 5 } & 0 \\ 0 & 0 & 5 \\ \end{array} \right) $$
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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 3 }{ 5 } & 1 & 0 \\ 0 & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 8 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 3 }{ 5 } & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & \frac{ 6 }{ 5 } & 0 \\ 0 & 0 & 5 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 3 }{ 5 } & 1 & 0 \\ \frac{ 3 }{ 5 } & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & \frac{ 6 }{ 5 } & 0 \\ 0 & 0 & 5 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 3 }{ 5 } & \frac{ 3 }{ 5 } \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 8 \\ \end{array} \right) $$