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Another geometry problem? Noooo.

Prove that a right angled triangle with all integer sides has an area that is divisible by 6.

Solution:

Let's look at the fundamental Pythagorean triplets instead of looking at their similar triangles. So we have the sides: $$(m^2+n^2), 2mn, (m^2-n^2)$$ , where m and n are two coprime natural numbers.

The area becomes: $$(m-n)(m+n)mn.$$ As m and n are coprime, if they are of the same parity, then they both have to be odd and hence, a factor of 4 and hence 2, is established in side $$m^2-n^2.$$ If the parity is even for either one of them, then a factor of 2 is established in the product mn. Now we need to look for a factor of 3 in order to prove the question.

If either one of m and n is a multiple of 3, then there's no question about it since the factor gets established in the product mn.

Otherwise, W.L.O.G*, let m and n have the configuration:

• 3k+1, 3l+1.
• 3k+1, 3l+2.
• 3k+2, 3k+2.

respectively

We can immediately see that in the first and last cases, the difference of m and n creates a factor of three in the area. In the second case, the sum creates a factor of three. Hence the area is going to be divisible by 6.

Q.E.D

*= Without Loss Of Generality.

If you want to impress your friends and relatives at a dinner party or whatever, you can ask them to do this:

• Think of any two numbers.
• Add those two numbers, subtract those two numbers and multiply the two numbers.
• Now write down your prediction on a piece of paper that says the product of the 3 resultant numbers is going to be a multiple of 3.
• They will think that you got lucky. So let them have a go at it again and again and again.... till they give up and ask you how to do it. Explain the method and look like a genius.

Note by Vishnu C
2 years ago