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Number theory Difficulty 5.4 AIME, harder Prove it Austria

Determine all nonnegative integers nn having two distinct positive divisors with the same distance from n2\frac{n}{2}.

Solution

Since the smallest possible divisors of an integer nn are 11, 22 and 33, the greatest possible divisors are nn, n2\frac{n}{2} and n2\frac{n}{2}. Hence a divisor that is bigger than n2\frac{n}{2} can only be nn or n2\frac{n}{2}. Since there is no positive divisor of nn having the same distance from n2\frac{n}{2} as nn, the bigger one of the two divisors must be n2\frac{n}{2}. The distance from n2\frac{n}{2} to n2\frac{n}{2} equals n2\frac{n}{2} and since n2n2=n2\frac{n}{2} - \frac{n}{2} = \frac{n}{2} the smaller divisor must be n2\frac{n}{2}. Hence nn is a multiple of 66. On the other hand it is clear that all positive multiples of 66 have the desired property.

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