Equation in terms of divisors.

Let n n be a natural number and 1 = d 1 < d 2 < . . . < d k = n 1=d_{1}<d_{2}<...<d_{k}=n be the positive divisors of n n .

Now,find the minimum n n such that 2 n = d 5 2 + d 6 2 1 2n=d_{5}^{2}+d_{6}^2-1


The answer is 272.

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