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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)

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  • $\begingroup$ Actually what was the exercise? $\endgroup$
    – Praveen
    Dec 28, 2014 at 17:07
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    $\begingroup$ Note that any square is congruent to $0$ or $1$ modulo $3$. So if neither $a$ nor $b$ is divisible by $3$, then $a^2+b^2\equiv 2\pmod{3}$, which is impossible for a square. $\endgroup$ Dec 28, 2014 at 17:09
  • $\begingroup$ @supremum: prove that in a pythagorean triple, a or b has to be divisible by 3. Should I put it in the body of the question? I thought it was a bit repetitive but maybe is confusing. $\endgroup$
    – Hugo
    Dec 28, 2014 at 18:38
  • $\begingroup$ @AndréNicolas: thanks! do you want to write that as a answer so I can mark it as accepted? $\endgroup$
    – Hugo
    Dec 28, 2014 at 18:39
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    $\begingroup$ You are welcome. When you find out how to do a problem, it is encouraged that you write it up yourself as an answer. If you have trouble with the LaTeX, just write it up as best you can and if you send me a message I can edit things. You made a good choice of book, Silverman's is well-written. $\endgroup$ Dec 28, 2014 at 19:01

3 Answers 3

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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.

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    $\begingroup$ The proof above shows that in any Pythagorean triple $(a,b,c)$, we have $3$ divides $a$ or $3$ divides $b$. This "or" as usual includes the possibility of both. For primitive triples, it cannot be both. $\endgroup$ Dec 28, 2014 at 21:06
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\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.

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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.

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  • $\begingroup$ I think there is a little mistake in the first simplified term (but it doesn't affect your conclusion), I think the first term should be (3x - 3p + y - q) $\endgroup$
    – Hugo
    Sep 10, 2015 at 12:26
  • $\begingroup$ @Hugo Aah yes. Thank you. $\endgroup$
    – the_kulk
    Sep 10, 2015 at 12:28

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