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in Olympiad by (65.8k points)

Find all natural number n such that 1 + [√2n] divides 2n. (For any real number x, [x] denotes the largest integer not exceeding x.)

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Now every solution of equation (3) will be the answer became here

2n = p(p +1) and 1+ [√2n] = p + 1 which divided p(p +1) = 2n

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