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Three consecutive integers form the lengths of a right-angled triangle. How many sets of such three consecutive integers is/are possible?

Three consecutive integers form the lengths of a right-angled triangle. How many sets of such three consecutive integers is/are possible? Correct Answer Only one

let n-1, n, n+1 be 3 consecutive integers

So

(n+1)^2= n^2+ (n-1)^2

(n+1)^2-(n-1)^2= n^2

4n = n^2

So n = 0 or n = 4

n can’t be 0 as n-1 will be negative then

So 3,4 and 5 is the only triplet formed.

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