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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.


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.


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.


*= 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
1 year, 5 months ago

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