# 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?

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Do you know modular arithmetic? The only squares modulo three are zero and one, and so the sums of two squares can only be 0 if both of the squares are congruent to 0. – anon Aug 26 '12 at 5:22

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.

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Well, we have two possibilities, first if neither $a_1$ nor $b_1$ is divisible by $3$ then that means that they have the form $a_1 = 3k \pm 1$ and $b_1 = 3t \pm 1$. Then $$a_1^2 + b_1^2 = (3k \pm 1)^2 + (3t \pm 1)^2 = 9k^2 \pm 6k + 1 + 9t^2 \pm 6t + 1 = 3A + 2$$

where $A = 3k^2 \pm 2k + 3t^2 \pm 2t$ and thus since $a_1^2 + b_1^2 = 3A + 2$ we can conclude that $3 \nmid a_1^2 + b_1^2$.

The other possibility is that exactly one of $a_1$ or $b_1$ is divisible by $3$. Let's assume that $a_1$ is divisible by $3$ but $b_1$ isn't. Then they have the form $a_1 = 3k$ and $b_1 = 3t \pm 1$. Thus

$$a_1^2 + b_1^2 = (3k)^2 + (3t \pm 1)^2 = 9k^2 + 9t^2 \pm 6t + 1 = 3B + 1$$

where $B = 3k^2 + 3t^2 \pm 2t$ so again since $a_1^2 + b_1^2 = 3B + 1$ then $3 \nmid a_1^2 + b_1^2$.

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