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?

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    $\begingroup$ Apart from the second, which is harder, they can be verified by looking at the parametric form of the general primitive triple. $\endgroup$ Sep 30, 2018 at 11:51
  • $\begingroup$ @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. $\endgroup$ Sep 30, 2018 at 12:52

1 Answer 1


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.


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