<s>[INST] Proof of "equations cannot have an integral solutions" using modular arithmetic. We a got an equation $$15x^2 -7y^2 =9$$ and we have to prove that this equation cannot have an integral solution. So, this what we will do:
$7y^2 = 15x^2 -9$, and since $3| 15x^2 -9 \implies 3|7y^2$. As 3 cannot divide 7 therefore, it must divide $y^2$ and hence it must divide $y$. So, let $y=3m$. Let's substitute this value of $y$ into our original equation:
$$
15x^2 - 63m^2 = 9 \\
5x^2 = 3+21m^2
$$
Since, 3 divides RHS, so it would divide LHS, that is $3|5x^2$, but 3 doesn't divide 5 so it must divide $x^2$ and hence it must divide $x$, so let $x= 3n$. Let's substitute this value in the last equation :
$$
45n^2 = 3+21m^2\\
15n^2= 1+7m^2
$$ 
That is, $ 7m^2 +1 \equiv 0 \mod 3$ but thats not possible because $$m \equiv 0 \implies m^2 \equiv 0 \\
m\equiv 1 \implies m^2 \equiv 1 \\
m \equiv 2 \implies m^2 \equiv 4 \implies m^2 \equiv 1 $$ (all mod 3)
And 
$$
7m^2 \equiv 0 ~~~~~~~~~~~~~~~~~~~~~~(i)\\
7m^2 \equiv 7 \implies 7m^2 \equiv 1~~~~~~~~~~~~~~~~~~~~~~~~(ii) $$
But we know $$
1 \equiv 1 \mod 3 \\
\text {Adding congruence relations of (i) and (ii) to the above one, we get}\\
7m^2 +1 \equiv 1 \\
7m^2 +1 \equiv 2 $$ 
Hence, $7m^2 +1 \equiv \mod 3$ is not possible. Therefore, the equation doesn't have any integral solutions (and I really don't know why does this show that our original equation doesn't have any integral solutions).

Let's consider this equation $$3x^3 +y^3 =6$$
We want to prove that this equation doesn't have any integral solution. So, we've $y^3 = 6-3x^3$, that means $3| y^3$ which implies $y= 3m$. Let's substitute this into our original equation, 
$$
3x^3 +27m^3 = 6 \\
x^3 + 9m^3 = 2\\
9m^3 = 2-x^3 
$$
That means, $2-x^3 \equiv 0 \mod 3$. So, we have 
$$
x \equiv 0 \implies -x^3 \equiv 0 \\
x \equiv 1 \implies -x^3 \equiv -1 \\
x \equiv 2 \implies x^3 \equiv 8 \implies x^3 \equiv 2 \implies -x^3 \equiv -2 ~~~~~~~~~~(iii)$$
(all are mod 3)
And we know 
$$
2 \equiv 2 \mod 3$$
Now, adding the congruence relation (iii) with the above one we get 
$$
2-x^3 \equiv 0 \mod 3$$ as wanted. But this doesn't prove that our equation have integral solutions (someone said that) and of course, our equation doesn't have any integral solutions but why we didn't reach any contradiction in the way we reached in the first proof. 
So, my question is why in the first proof when reached a contradiction we concluded "no integral solutions are possible" but in the second proof everything agreed with one-another but we didn't conclude "integral solutions are possible", Why? 
Please explain me. 
 [/INST] Given
$$15x^2 -7y^2 =9\quad\implies\quad x = ± \frac{\sqrt{7 y^2 + 9}}{\sqrt{15}}\quad\land\quad y = \sqrt{\frac{3}{7}} \sqrt{5 x^2 - 3}$$
we can see that there are no rational solutions,  let alone integers.
</s>