is it must that a Pythagoras triplet should have one of its number as a multiple of 3
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No definately no it is false
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considered the pythagoras theoram :
a²+ b² + c² = c²
let a² = 3m or a²= 3m +1 ( a prefect square can be expressed as a multiple of 3.
similarity, let b²= 3k or b²= 3k + 1
then c² according to the pythagoras theoram, the sum of a² and b².
let us assume that we take a²and b² as non - multiple of 3.
c²=3m +1 +3k+1= 3(m+k)+ 2
put simplicity the successor of the successor of a multiple of 3. This is a contradiction since c² is a perfect square which can only be expressed as a a multiple of 3 or the successor of 3.
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