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I'm reading "Friendly Introduction to Number Theory". Now I'm working on Primitive Pythagorean Triples Exercises 2.1 (a) on P18.

We showed that in any primitive Pythagorean triple $(a, b, c)$, either $a$ or $b$ is even. Use the same sort of argument to show that either $a$ or $b$ must be a multiple of 3.

(1) $a^2 + b^2 = c^2$ with a odd, b even, a,b,c having no common factors

(2) $a^2 = c^2 - b^2 = (c-b)(c+b)$

(3) $c + b = s^2$ and $c - b = t^2$

(4) $c = \frac{(s^2 + t^2)}{2}$ and $b = \frac{(s^2 - t^2)}{2}$

(5) $a = \sqrt{(c-b)(c+b)} = st$

(6) $a = st$, $b = \frac{(s^2 - t^2)}{2}$, $c = \frac{(s^2 + t^2)}{2}$


I have no idea how I start doing this. Can you give me a hint? I think I need to show that both the following (1) and (2) are satisfied.

$X \neq 0$

(1) $a\equiv 0\pmod 3$ and $b\equiv X\pmod 3$

(2) $b\equiv 0\pmod 3$ and $a\equiv X\pmod 3$


marked as duplicate by Bill Dubuque divisibility Oct 18 '18 at 15:54

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The numbers $a$ and $b$ cannot both be multiples of $3$, because then $c$ would be a multiple of $3$ and the triple would not be primitive.

And if none of them is a multiple of $3$, then both of them are of the form $3k\pm1$, for some integer $k$, from which it follows that both squares $a^2$ and $b^2$ are of the form $3k+1$. But then $c^2(=a^2+b^2)$ is of the form $3k+2$. You should be able to show that this is impossible.

  • $\begingroup$ ooh! I should have used proof by contradiction. I understand the first one (both be multiples of 3) . Now still I'm still thinking about the second one why the form is 3k±1.. $\endgroup$ – zono Oct 18 '18 at 9:59
  • $\begingroup$ If $a$ is not a multiple of $3$ then, since the $3$ numbers $a-1$, $a$, and $a+1$ are consecutive and therefore one of them is a multiple of $3$, then $a-1$ or $a+1$ is a multiple of $3$. $\endgroup$ – José Carlos Santos Oct 18 '18 at 10:05
  • $\begingroup$ Thank you! I understand it. If possible.. can you tell me a little bit more about "3k+2". Although I took 15 min to think about it but I could not get it.. "c^2(=a^2+b^2) is of the form 3k+2" $\endgroup$ – zono Oct 18 '18 at 10:24
  • $\begingroup$ You don't understand why is it that when I add two numers which are a multiple of $3$ plus $1$, then what I get is a multiple of $3$ plus $2$? $\endgroup$ – José Carlos Santos Oct 18 '18 at 10:26
  • 1
    $\begingroup$ @zono: We have $(3k+1)^2=3\cdot (3k^2+2k)+1$ and $(3k-1)^2=3\cdot (3k^2-2k)+1$; the sum of two such squares would be $3\cdot(\text{something})+1+1$. $\endgroup$ – Henning Makholm Oct 18 '18 at 12:55

Using https://en.m.wikipedia.org/wiki/Pythagorean_triple or https://mathcs.clarku.edu/~djoyce/java/elements/bookX/propX29.html

$a=2mn,b=m^2-n^2$ where $m,n$ are coprime integers and not both are odd


Now use The product of $n$ consecutive integers is divisible by $n$ factorial as $m^3-m=(m-1)m(m+1)$


In mod 3, $0^2=0$, $1^2=1$ and $2^2=(-1)^2=1$. Thus, if neither $a$ nor $b$ are 0 mod 3, then $a^2+b^2$ is 2 mod 3, but no number squares to 2 mod 3.

The "principal" condition doesn't really play a part in the proof.

There are also proofs that at least one number must be divisible by 4, and that at least one number must be divisible by 5.


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