$\text{mod 3}$ gives $3|a$ and $3|c$. Writing $a=3a_1$ and $c=3c_1$ gives $5a_1^2+b^2=13c_1^2$ and $\text{mod 5}$ finishes because of infinite descent. Only solution is $(0, 0, 0)$ @below, I just realized $\text{mod 3}$ was nonsense, thanks.

rafaello wrote: Find all integer triples $(a, b, c)$ satisfying the equation $$ 5 a^2 + 9 b^2 = 13 c^2. $$ $\pmod{5}$ gives $5\mid b,c$. Define $b=5b',c=5c'$. We have $a^2+45b'^2=65c'^2$ and $5\mid a$. Define $a=5a'$. We have $5a'^2+9b'^2=13c'^2$. By infinite descent we get $a=b=c=0$.

rafaello wrote: Find all integer triples $(a, b, c)$ satisfying the equation $$ 5 a^2 + 9 b^2 = 13 c^2. $$ Too simple by Quadratic Reciprocity - Notice $\mod 13$ , we have $$ 5a^{2} + 9b^{2} \equiv 0 \mod 13$$, giving $$5a^{2} \equiv 4b^{2} \mod 13$$, but note that as $4b^{2}$ is a $QR$ $\mod 13$ , giving $\left ( \frac{5}{13} \right ) = 1$ , but from Quadratic reciprocity $$ \left ( \frac{5}{13} \right ).\left ( \frac{13}{5} \right ) = (-1)^{even} = 1$$, but as $\left ( \frac{13}{5} \right ) = -1$ , this is a contradiction . Hence only solution is $(0,0,0)$. $\blacksquare$.