How to prove that
No solutions exist for $2a^2 + 3b^2 = 6c^2$ for $a,b,c \in N$
Attempt:
Since $3∣3b^2$ and $3∣6c^2$ , we must have $3∣2a^2$ and thus $3∣a$.
Since $2∣2a^2$ and $2∣6c^2$ , we must have $2∣3a^2$ and thus $2∣b$.
Let $a = 3A , b = 2B$
$18A^2 + 12B^2 = 6c^2$
$3A^2 + 2B^2 = c^2$
I am stuck here. Reducing this equation using modular arithmetic does not seem to help.
