Quote:
Originally Posted by sjastro
Now the question is if there is a case where "a" is an even integer for the Liethagoras condition to be true?
Regards
Steven
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Not possible.
If a = 2n then a^2 = 4n^2, which is also even.
However, c-b = 1 implies one of b or c is even, the other odd.
Therefore b+c is odd.
i.e. a^2 cannot equal b+c as LHS is even but RHS is odd.
Regards, Rob.