Sum of consecutive numbers

Let n n be a positive integer, such that there exist an other positive integer k k ( k > n k>n ), where 1 + 2 + 3 + + n = ( n + 1 ) + ( n + 2 ) + ( n + 3 ) + + k {\color{#3D99F6}{1+2+3+\dots+n}}={\color{#D61F06}{(n+1)+(n+2)+(n+3)+\dots+k}} If the third smallest possibe value of n n is N N , and the thrird smallest possible value of k k is K K , then find the value of ( K N ) 2 (K-N)^2


The answer is 1225.

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