proof: primitive pythagorean triple, a or b has to be divisible by 3

Question

I'm reading "A friendly introduction to number theory" and I'm stuck in this exercise, I'm mentioning this because what I need is a basic answer, all I know about primitive pythagorean triplets is they satisfy $a^2 + b^2 = c^2$ and a, b and c has no common factors.

Now.. my approach (probably kind of silly) was to "classify" the odd numbers (not divisible by 3) as $6k+1$, $6k+2$ and $6k+5$, and the even numbers with $6k+2$ and $6k+4$, then, trying different combinations of that, I could probe all the cases when I assume c is not 3 divisible, but I still have to probe that c cannot be 3 divisible and I don't know how to do it.

Anyway, probably there is a better simpler solution.

(Sorry, if this is a stupid question, I'm trying to teach myself number theory without much math background)

Answer 1

Any square is congruent to $0$ or $1$ modulo $3$

So having, $a^2 + b^2 = c^2$

Let's suppose neither $a$ nor $b$ is divisible by $3$, then, the squares must be $1$ modulo $3$.

So, the expression can be re-written as:

$(3k + 1) + (3k' + 1) = c^2$

and then

$3 (k + k') + 2 = c^2$

That is, $c^2$ is a square congruent $2$ modulo $3$, which is absurd.

Edit: maybe I should add that for definition of Pythagorean triple, only one can be divisible by 3, not both.

Answer 2

\begin{array}{|c|c|} \hline n \pmod 3 & n^2 \pmod 3 \\ \hline 0 & 0 \\ 1 & 1 \\ 2 & 1 \\ \hline \end{array}

If neither $a$ nor $b$ is a multiple of $3$, then $a^2 + b^2 \equiv c^2 \pmod 3$ becomes $1 + 1 \equiv c^2 \pmod 3$, which simplifies to $c^2 \equiv 2 \pmod 3$; which has no solution.

Answer 3

I'm reading the book too. I came up with a different approach. Of course, it is not as elegant as the answer given, but here goes.

$$ a = st $$ $$ b = (s^2 - t^2)/2 $$

Choose values for s and t.

If either of s or t is a multiple of 3, then $ a = st $ is a multiple of 3. Both s and t cannot be multiples of 3, as that will cause both a and b to be multiples of 3 and the triplet won't be primitive.

If neither s nor t is a multiple of 3.

$$ s = 3x + y $$ $$ t = 3p + q $$

Here y and q can take values 1 and 2.

a cannot be a multiple of 3. Simplifying b, we get

$$ b = (s - t)(s + t)/2 $$ $$ b = (3x - 3p + y - q)(3x + 3p + y + q)/2 $$

Now y and q can either be 1 or 2.

When y and q are both equal to 1 or 2, the first term becomes $(3x - 3p)$.

When y and q are unequal, but take the values 1 or 2, the second term becomes $(3x + 3p + 3)$.

In both cases, b is a multiple of 3.