# Pythagorean triples

• Jan 27th 2010, 11:34 AM
zhupolongjoe
Pythagorean triples
Show that the area A of a triangle whose sides form a pythagorean triple is always divisible by 6.

Hint: For an integer x, x^2 is of the form 4k or 1+8k, and when 3 doesn't divide x, x^2 is of the form 1+3k.

I don't know what to do. I figured the strategy would be to show that it is divisible by 2 and also by 3, so that it is divisible by 6=2*3, but I don't know how to go about this or if it is a correct approach.

thanks
• Jan 27th 2010, 12:51 PM
Dinkydoe
I hope you can do a little modular arithmatic. You're idea is indeed the right one:

Let $\displaystyle (a,b,c)$ be such that $\displaystyle a^2+b^2= c^2$
The area of the triangle is given by: $\displaystyle A = \frac{1}{2}ab$

Observe that: $\displaystyle c^2\equiv 0,1$ mod 4. And $\displaystyle c^2\equiv 0,1$ mod 3. (you can easily check this by looking at quadratic residu-classes modulo 4, and modulo 3).

Case 1.
Assume $\displaystyle c^2=a^2+b^2\equiv 0$ mod $\displaystyle 4 \Longrightarrow a^2\equiv b^2\equiv 0$ mod 4 $\displaystyle \Longrightarrow ab\equiv 0$ mod 4.

Assume $\displaystyle c^2= a^2+b^2\equiv 0$ mod 3$\displaystyle \Longrightarrow a^2\equiv b^2\equiv 0$ mod 3$\displaystyle \Rightarrow ab\equiv 0$ mod 3.

Case 2.
Assume $\displaystyle c^2 = a^2+b^2\equiv 1$mod 4. This implies $\displaystyle a^2+b^2\equiv 1$ mod 8, or $\displaystyle a^2+b^2\equiv 5$ mod 8

Now we'll look at quadratic residues modulo 8 wich are $\displaystyle \left\{0,1,4\right\}$

Since there is no quadratic residu that is 5 mod 8. It follows that $\displaystyle c^2=a^2+b^2\equiv 1$ mod 8$\displaystyle \Rightarrow a^2\equiv 0$ mod 8, $\displaystyle b^2\equiv 1$ mod 8. Hence $\displaystyle ab\equiv 0$ mod 8.

Assume $\displaystyle c^2= a^2+b^2\equiv 1$ mod 3 $\displaystyle \Longrightarrow a^2\equiv 0$ mod 3, $\displaystyle b^2\equiv 1$ mod 3 $\displaystyle \Rightarrow ab\equiv 0$ mod 3

So in both cases we have $\displaystyle 4|ab$ and $\displaystyle 3|ab\Rightarrow 6|A$
• Jan 27th 2010, 01:54 PM
zhupolongjoe
Thanks so much!
• Jan 27th 2010, 02:54 PM
Dinkydoe

I argued that if $\displaystyle c^2 = a^2+b^2\equiv 1$ mod 4 that the only possiblility is that $\displaystyle c^2=a^2+b^2\equiv 1$ mod 8

This is an important observation!
• Jan 29th 2010, 02:18 PM
zhupolongjoe
I'm sorry, I have one more question. I understand your proof, but wouldn't it be easier to do it with mod 3 and mod 2 instead of mod 3 and mod 4, or is there some reason why you can't consider it with mod 2?

Thanks.
• Jan 29th 2010, 03:25 PM
Dinkydoe
We want to proof that $\displaystyle 2|(\frac{ab}{2})$ hence we must prove $\displaystyle 4|ab$. That's why $\displaystyle ab\equiv 0$ mod 2 isn't enough.
• Jan 29th 2010, 03:31 PM
zhupolongjoe
Sorry for all of the questions, but then wouldn't we need to show that 6 divides ab instead of 3 since we need to show that 3 divides (ab)/2.
• Jan 30th 2010, 03:43 AM
Dinkydoe
Quote:

Sorry for all of the questions, but then wouldn't we need to show that 6 divides ab instead of 3 since we need to show that 3 divides (ab)/2
No. if $\displaystyle 3|ab$ then it follows $\displaystyle 3|\frac{ab}{2}$, since 2,3 are coprime. Do I need to proof this for you? ;p

If we show that $\displaystyle 3|ab$ and $\displaystyle 4|ab$ then we've shown that $\displaystyle 12|ab$. Hence $\displaystyle 6|{\frac{ab}{2}}$.
• Jan 30th 2010, 06:21 AM
zhupolongjoe
Ok, I've got it now. Thanks for your patience.