# Proof for General Properties of Pythagorean Triples

Having read the Wikipedia article "Pythagorean Triple", I came across the "Elementary properties of primitive Pythagorean triples" section which described many conditions for primitive triples, namely:

Exactly one of a, b is odd; c is odd.

At most one of a,b,c is a square

Exactly one of a,b is divisible by 3

Exactly one of a,b is divisible by 4

Exactly one of a,b,c is divisible by 5

I would like to know if there are proofs that exist for these statements. If so, would anyone be kind enough to (mathematically) explain to me why these statements are true?

• Apart from the second, which is harder, they can be verified by looking at the parametric form of the general primitive triple. Sep 30, 2018 at 11:51
• @lord-shark-the-unknown Would you mind sharing how that can be done? As someone with exceedingly more experience in mathematics than me, it must seem obvious. Unfortunately, I can't deduce the truth of the statements above just by inspection. Sep 30, 2018 at 12:52

With regard to factors of $$2$$ (odd/even), $$3,4,5$$, it should be clear that at most one of $$a,b,c$$ can have each of those as a factor, since if two of them had a common factor, the third would also have to have that factor and the triple would not be primitive.
Regarding $$2$$: If $$c$$ is even, then both $$a,b$$ must be odd. In that case, $$a\equiv \pm 1 \mod{4}$$ and $$b\equiv \pm 1 \mod{4}$$. Hence $$a^2\equiv 1 \mod{4}$$ and $$b^2\equiv 1\mod{4}$$, so $$c^2\equiv 2\mod{4}$$. Even numbers that are congruent to $$2$$ $$\mod{4}$$ (such as $$2,6,10,14,18$$, etc.) are simply twice an odd number. They contain only one factor of $$2$$, and hence cannot be perfect squares. So $$c$$ cannot be even, and must be odd. This means that both $$a,b$$ cannot be odd (or the sum of their squares would be even), so one must be even.
Regarding $$3$$: Any number $$\mod{3}$$ must be congruent to one of $$-1,0,1$$. So the square of any number $$\mod{3}$$ must be congruent to either $$0,1$$. The only way you can make a sum involving three squares $$\mod{3}$$ is $$1+0=1$$. So one of $$a,b$$ must be divisible by $$3$$.
Regarding $$4$$: This result is most easily understood from considering the generating formula for Pythagorean triples. All primitive Pythagorean triples can be generated from the relationships $$a=m^2-n^2, b=2mn, c=m^2+n^2$$ where $$m>n$$ are integers. It is apparent that $$b$$ is the even member of the triple, and for $$a,c$$ to be odd, it is apparent that $$m,n$$ must have different parity. In other words, one of $$m,n$$ must be even. Hence $$2mn$$ must be divisible by $$4$$.
Regarding $$5$$: Any number $$\mod{5}$$ must be congruent to one of $$0,\pm 1,\pm 2$$. So the square of any number $$\mod{5}$$ must be congruent to one of $$0,\pm 1$$. You can't make a sum involving three such numbers which omits $$0$$, that is, consists of only $$1$$ and $$-1$$. So one of the numbers must be divisible by $$5$$.
The final fact (at most one of $$a,b,c$$ can be a square) is demonstrated in the proof of Fermat's Last Theorem ($$x^n+y^n=z^n$$ has no solutions for $$n>2$$) for the specific exponent $$n=4$$. Beyond the scope of this answer.