# For any primitive pythagorean triple $(a,b,c)$ either $a$ or $b$ must be a multiple of $3$ [duplicate]

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

https://www.math.brown.edu/~jhs/frintch1ch6.pdf

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 $$a\equiv X\pmod 3$$

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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.

• 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.. – zono Oct 18 '18 at 9:59
• 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 $3$. – José Carlos Santos Oct 18 '18 at 10:05
• 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" – zono Oct 18 '18 at 10:24
• 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 $2$? – José Carlos Santos Oct 18 '18 at 10:26
• @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 $3\cdot(\text{something})+1+1$. – Henning Makholm Oct 18 '18 at 12:55

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

$$ab=2mn(m^2-n^2)=2(m^3-m)n-2m(n^3-n)$$

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.