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 $F(a,b,c)\equiv 0 \pmod{13^2}$. 
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.
 A: 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 $169$.
