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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?

Correct Answer: Only one

Explanation:

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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