Infinite descent This Wikipedia article of Infinite Descent says:

We have
  $ 3 \mid a_1^2+b_1^2 \,$.
  This is only true if both $a_1$ and $b_1$ are divisible by $3$.

But how can this be proved?

Suppose $a_1 = 3 q_1 + r_1$ and $b_1 = 3 q_2 + r_2$, where $r_1$ and $r_2$ is either $-1$, $0$ or $1$. Then
$$
  a_1^2 + b_1^2 = 3 \left( 3 q_1^2 + 3 q_2^2 + 2 q_1 r_1 + 2 q_2 r_2 \right) + r_1^2 + r_2^2
$$
For $a_1^2 + b_1^2$ to be divisible by $3$, we should have $r_1^2 + r_2^2 = 0$, since $0\leqslant r_1^2+r_2^2 < 3$.  Enumerating 9 cases, only $r_1 = r_2 = 0$ assure the divisibility.
This is essentially the answer anon gave in his comment.
