Solve the following number theory problem with 2 variables

Question

Let there be $$a,b∈ \Bbb Z$$ Demonstrate that there exist no solutions for the following equation $$a^2-3b^2=-1$$

Answer 1

As per the comment by Crostul,$$a^2-3b^2=-1 \Rightarrow a^2+1=3b^2$$

Now $3$ divides R.H.S.
But
for any $a$, $$a \equiv 0,1,2 \pmod 3 \Rightarrow a^2 \equiv 0, 1 \pmod 3 \Rightarrow a^2+1 \equiv 1, 2 \pmod 3$$

So $3$ does not divide L.H.S.

Hence, no solution.

Answer 2

Let's assume a and b are both even. Then $ { 4k }^{ 2 } = 3(4q^2)-1$ and so $2(2k^2)=2(6q^2)-1$ and... $2(x)=2(y)-1$ odd can't be even. Let's assume now a and be are both odd. Then $(2k+1)^2=3(2q+1)^2 -1$ So $2(2(k^2+k)) +1 = 2(6q^2+6q+1)$ and... $2(x) +1 = 2(y)$ even can't be odd

The last one I let you solve it on your own.