# Observations on integer-sided right triangles

Prove that in any integer side right angled triangle the following hold true:
one side is always divisible by 3
one side is always divisible by 5
the product of two legs is always divisible by 12
the product of all three sides is always divisible by 60
I have observed the above properties in all the integer right angled triangles and iam looking for a proof here legs in the sense:the two sides other than the hypotenuse

• You should change and say that there is at least two legs whom product is divisble by $12$. Phrased like you did, it seems like any two leg give a product divisible by $12$, which is false. – Martigan Feb 18 '15 at 12:19
• @Mar, a right triangle only has two legs --- the hypotenuse is usually not called a leg. – Gerry Myerson Feb 18 '15 at 12:21
• Are you aware, lokesh, of the formula that gives all integer sided right triangles? – Gerry Myerson Feb 18 '15 at 12:22
• Everything you ask for is there: en.wikipedia.org/wiki/Pythagorean_triple – Martigan Feb 18 '15 at 12:22
• The formula $2ab,b^2-a^2,b^2+a^2$ includes $56,33,65$ and $20,21,29$ – Empy2 Feb 18 '15 at 14:02

The key point here is the fact that all Pythagorean triples can be generated as

$$a = 2kmn \qquad b = k(m^2-n^2) \qquad c = k(m^2+n^2)$$

For the sake of the following considerations, you may as well assume $k=1$, since if any of your claims holds for that case, it will automatically hold for $k>1$ as well.

The key point to all my answers is to consider this formula in modular arithmetic, so that you only need to consider a finite number of cases.

the product of two legs is always divisible by 12

$2mn(m^2-n^2)=2mn(m-n)(m+n)$ so you want to show that $mn(m-n)(m+n)\equiv0\pmod6$. Which you can break down to one consideration modulo 2 and one modulo 3. Modulo 2 is easy: If either $m$ or $n$ is even, you get $mn\equiv 0\pmod2$. If both are odd, you get $(m-n)\equiv0\pmod2$. If either $m$ or $n$ is divisible by $3$, you have $mn\equiv0\pmod3$. Otherwise you either have $m\equiv n$ and hence $m-n\equiv0$, or you have $m\equiv-n$ and hence $m+n\equiv 0$. This can be readily seen if you consider $\{-1,0,1\}$ as the set of representatives for integers modulo $3$.

the product of all three sides is always divisible by 60

$2mn(m^2-n^2)(m^2+n^2)=2mn(m-n)(m+n)(m^2+n^2)$ so you want to verify $mn(m-n)(m+n)(m^2+n^2)\equiv0\pmod{30}$. Which means it has to be zero modulo 2, modulo 3 and modulo 5. Modulo 2 and 3 we did above, multiplying one more term will not change that. So consider modulo 5. If either $m$ or $n$ is divisible by $5$ you're done. If $m\equiv n$ or $m\equiv-n$ the corresponding factor will be zero. So what cases remain? One of $m$ or $n$ must be $\pm1$, the other must be $\pm2$, modulo 5. Which means $m^2+n^2\equiv(\pm1)^2+(\pm2)^2=5\equiv0$.

one side is always divisible by 5

If the product of the three sides is divisible by 5, then one side has to be divisible by 5, because that factor has to come from one of the sides.

one side is always divisible by 3

Likewise: if the product of the sides is divisible by 3, then one side has to be. As an alternative, you could consider the possible edges modulo 3 and simply tabulate them:

$$\begin{array}{c|ccccccccc} m & 0 & 0 & 0 & 1 & 1 & 1 & 2 & 2 & 2 \\ n & 0 & 1 & 2 & 0 & 1 & 2 & 0 & 1 & 2 \\\hline a & 0 & 0 & 0 & 0 & 2 & 1 & 0 & 1 & 2 \\ b & 0 & 2 & 2 & 1 & 0 & 0 & 1 & 0 & 0 \\ c & 0 & 1 & 1 & 1 & 2 & 2 & 1 & 2 & 2 \end{array}$$

As you can see, there is always at least one zero in the bottom three rows, so you have at least one edge divisible by three. Even more, there is always one zero in the rows $a$ and $b$, so you always get a leg divisible by three. Not surprising, since the product of the two legs already contained that factor 3.

This tabulation approach would work for all the other answers as well, but some of the tables would be really big (like $60^2$ columns if you do it the naive way).

• The product of all three sides is divisible by $5$ (from the second fact) and also by $12$ (from the third fact), from which I think the fourth fact follows immediately. – David K Feb 18 '15 at 13:25
• No, there are many many many primitive Pythagorean triples that don't come from those formulas. You can't get $5,12,13$ that way, for instance. – Gerry Myerson Feb 19 '15 at 6:02
• @GerryMyerson: You are right, although $k=-3$ comes close. I guess I'm too used to working with homogeneous coordinates, where rational and integral are essentially the same thing. So the OP's comment claiming this rang so true that I didn't question it. Should have, in hindsight. I'm curious whether I'll be allowed to delete an accepted answer, until I find a way (and the time) to fix this… – MvG Feb 19 '15 at 7:55
• OK, fixed my answer. – MvG Feb 19 '15 at 8:51