# Show that there are infinitely many triples of integers $(a,b,c)$ such that $2a^2 + 3b^2 - 5c^2 = 1997$

(Cono Sur Math Olympiad - 1997) Show that there are infinitely many triples of integers $(a,b,c)$ such that $2a^2 + 3b^2 - 5c^2 = 1997$.

I tried to attribute a value to $a$ or $b$ to put this equation in the Pell's form, but I hadn't success.

I expect what they had in mind was this: as $1997 \equiv 5 \pmod {24}$ and is prime, it can be written as $1997 = 2 \cdot 31^2 + 3 \cdot 5^2.$ Note that the form $f(x,y)= 2 x^2 + 3 y^2$ represents all primes $p=2,3$ and all $p \equiv 5,11 \pmod {24}.$
So the first solution could be $$(a,b,c) = (31,5,0).$$ The automorphism group of $2 a^2 - 5 c^2$ is not difficult. We actually can find all solutions to $2 a^2 - 5 c^2 = 1922,$ but we don't need to. For any solution $$(a,b,c)$$ to the original problem, there is a new one $$(19a + 30 c, b, 12 a + 19 c).$$ So, an infinite sequence of solutions is $$(31,5,0),$$ $$(589,5,372),$$ $$(22351,5,14136),$$ $$(848749,5,536796),$$ and so on.
A different sequence is $$(5,28,9),$$ $$(365,28,231),$$ $$(13865,28,8769),$$ $$(526505,28,332991),$$ and so on.
We can instead vary the $3 b^2 - 5 c^2.$ For any solution $$(a,b,c)$$ to the original problem, there is a new one $$(a,4 b + 5 c, 3 b + 4 c).$$ Mixing together the two Pell type transformations gives us a big mess'o solutions.