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No. Here's why: you have a² + b² = c². If a is odd, then a² will also be odd (and if a is even, then a² will also be even. Same goes true for b² and c².

If a² & b² are both odd, then you have odd + odd, which is even. Why is {odd + odd} always even, you may ask. An informal proof could be something like this:

  • Let m & n be non-zero integers.
  • So 2m and 2n are guaranteed to be even,
  • then (2m + 1) & (2n + 1) are both odd.
  • Add them together: (2m + 1) + (2n + 1) = 2m + 2n + 2 = 2(m + n + 1) which is even.

And why is a² odd, when a is odd: if a is odd, then it does not have 2 as one of its factors. Since you square the number, you have not added any additional factors, so it still does not have 2 as a factor. If a was even, then it will have 2 as a factor, when you square it, you now have 2 as well as 2² as a factor, so the square is even as well.

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Q: Can a right triangle have all side that are an odd number?
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