Show that this polynomial in two variables has no integer solutions
Hint : Considering in mod $m$ for some $m\ge 2\in\mathbb N$ sometimes works. In our case, $m=3$.
Added : It seems that you are not familiar with considering in mod $m$.
The equation can be written as $$y^2=3(5x^2-2y^2-6)+2$$ So, $y^2$ has to be of the form $3k+2$ where $k\in\mathbb Z$. This is impossible. (why?)
Consider the equation modulo 3: $$15x^2-7y^2=16$$
- $15x^2$ is always divisible by 3,
- $7y^2$ is either 0 or 1 modulo 3 and
- $16\equiv1\pmod3$.
Substituting these residues into the equation, we find that it is inconsistent in all cases: $$0-0\not\equiv1\pmod3$$ $$0-1\not\equiv1\pmod3$$ $$15x^2-7y^2\not\equiv16\pmod3$$ Therefore the equation has no integer solution.
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set mode to Step-by-Step.
First of all we must determine the $gcd$ of all coefficients but the constant term, that is: $gcd(15, 0, -7, 0, 0) = 1.$
Dividing the equation by the greatest common divisor we obtain: $15 x^2 - 7 y^2 - 16 = 0$.
We try to check the equation modulo the prime divisors of $15$ ... here the hint ends :) ...