# At least one member of a pythagorean triple is even

I am required to prove that if $a$, $b$, and $c$ are integers such that $a^2 + b^2 = c^2$, then at least one of $a$ and $b$ is even. A hint has been provided to use contradiction.

I reasoned as follows, but drew a blank in no time:

Let us instead assume that both $a$ and $b$ are odd. This means that $a^2$ and $b^2$ are odd, which means that $c^2$ is even. Thus, $c$ is even. But this is not a contradiction -- it's the sum of two odd numbers, after all.

• You can say even more about the squares of odd numbers than only that they are odd. And you can say more about the squares of even numbers than just that they are even. Think $4$. – Daniel Fischer Jul 23 '13 at 3:32
• You've considered the situation mod 2. How about looking at it mod 4? – AGM Jul 23 '13 at 3:33
• Thanks @DanielFischer and @AGM! I think I got it: If $c$ is even, then $c^2$ is divisible by 4. This means that both $a^2$ and $b^2$ are divisible by 4, which is a genuine contradiction! – dotslash Jul 23 '13 at 3:36
• If $a$ and $b$ are both odd, you've surmised that $c^2$ must be even. $a = 2k+1$ and $b = 2l+1$ for some integers $k,l$. So then $a^2 = 4k^2 + 4k + 1$ and $b^2 = 4l^2 + 4l + 1$. And so $a^2 + b^2 = 4(k^2 + k + l^2 + l) + 2$. Can you take it from here? – Cameron Williams Jul 23 '13 at 3:36
• @CameronWilliams I did so much in one of my early attempts, but then had no idea what to do next. – dotslash Jul 23 '13 at 3:37

Suppose that $a$ and $b$ are both odd, and proceed as you did to conclude that $c$ is even. As $a$ is odd, we may write $a = 2k + 1$ for some $k$; similarly, $b = 2m + 1$. As $c$ is even, write $c = 2n$. This leads to

$$(2k + 1)^2 + (2m + 1)^2 = (2n)^2$$

or upon expanding and regrouping,

$$4(k^2 + k + m^2 + m) + 2 = 4n^2$$

Now the right side is divisible by 4, as is the first term on the left - but $2$ is not divisible by $4$. Do you now see how to derive a contradiction?

• Yes, thanks a lot! – dotslash Jul 23 '13 at 4:25

Let us try to write this using congruences, which is a very useful and compact notation.

We know that a square can only have remainder $0$ or $1$ modulo $4$. I.e., for any integer $x$ one of these two congruences must be true: $x^2\equiv 0 \pmod 4$ or $x^2\equiv1 \pmod4$. The first case happens if $x$ is even, the second case if $x$ is odd. (Since $(2k+1)^2=4(k^2+k)+1$ and $(2k)^2=4k^2$.)

So we have two possibilities for the remainder of $a^2$ and two possibilities for the remainder of $b^2$ modulo $4$. $$a^2 \equiv 0 \pmod4, b^2 \equiv 0 \pmod4 \Rightarrow c^2\equiv 0+0=0 \pmod4\\ a^2 \equiv 1 \pmod4, b^2 \equiv 0 \pmod4 \Rightarrow c^2\equiv 1+0=1 \pmod4\\ a^2 \equiv 0 \pmod4, b^2 \equiv 1 \pmod4 \Rightarrow c^2\equiv 0+1=0 \pmod4\\ a^2 \equiv 1 \pmod4, b^2 \equiv 1 \pmod4 \Rightarrow c^2\equiv 1+1=2 \pmod4$$ We see that in the last case we would have $c^2\equiv 2\pmod4$, which is not possible.

So only the first three cases can really occur. In the other words, either all of the numbers $a$, $b$, $c$ are even, or exactly one of the is even and the remaining two numbers are odd.

• Great answer, Martin! The idea of divisibility by 4 was a revelation to me, really. Your approach is a bit beyond me as of now, as I'm not yet 100% comfortable with modular arithmetic. That said, it indeed is compact and useful. – dotslash Jul 23 '13 at 6:50