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If $(a,b,c)$ is a primitive Pythagorean triplet, explain why only one of $a$,$b$ and $c$ can be even-and that $c$ cannot be the one that is even.

What I Know:

A Primitive Pythagorean Triple is a Pythagorean triple $a$,$b$,$c$ with the constraint that $\gcd(a,b)=1$, which implies $\gcd(a,c)=1$ and $\gcd(b,c)=1$. Example: $a=3$,$b=4$,$c=5$ where, $9+16=25$

At least one leg of a primitive Pythagorean triple is odd since if $a$,$b$ are both even then $\gcd(a,b)>1$

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  • $\begingroup$ Doesn't Euler's formula $\{a,b,c\}=\{2mn,m^2-n^2,m^2+n^2\}$ (though $a,b,c$ are not necessarily PPTs) explain this? $\endgroup$ – 000 Dec 9 '12 at 0:46
  • $\begingroup$ @Limitless The problem is that this result is sometimes used in the proof of the Euler formula. $\endgroup$ – N. S. Dec 9 '12 at 0:48
  • $\begingroup$ @N.S. Ah. I was not aware. $\endgroup$ – 000 Dec 9 '12 at 0:48
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Clearly not all are even as then $\,(a,b,c)\,$ have a common divisor and this is not a primitive triple.

If $\,c\,$ as even then so would also be $\,a^2+b^2\,$, which means$\,a,b\,$ have the same parity and thus they're odd, but this can't be since then

$$a^2\,,\,b^2=1\pmod 4\Longrightarrow a^2+b^2=2\pmod 4=c^2$$

and since it $\,2\,$ is not a square modulo $\,4\,$ we're done.

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Hint $a^2,b^2,c^2$ are either $0$ or $1 \pmod 4$.

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Clearly they cannot all be even as a smaller similar triple could be obtained by dividing all the sides by $2$ (your final point).

Nor can two of them be even since $a^2+b^2=c^2$ and either you would have an even number plus an odd number (or the other way round) adding to make an even number or you would have an even number plus an even number adding to make an odd number.

Nor can they all be odd since $a^2+b^2=c^2$ and you would have an odd number plus an odd number adding to make an odd number.

Added: Nor can the two shorter sides be both odd, say $2n+1$ and $2m+1$ for some integers $n$ and $m$, as the longer side would be even, say $2k$ for some integer $k$, as its square is the sum of two odd numbers, but you would then have $4n^2+4n+1 + 4m^2+4m+1 = 4k^2$ which would lead to $k^2 = n^2+n+m^2+m + \frac{1}{2}$ preventing $k$ from being an integer.

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  • $\begingroup$ You have not explained why $a,b$ can't be odd and $c$ even, which is probably the hardest point of the question. $\endgroup$ – Ross Millikan Dec 9 '12 at 1:59
  • $\begingroup$ @Ross - true - I will edit with the standard mod 4 $\endgroup$ – Henry Dec 9 '12 at 8:42

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