## Integer partition equations

### Problem 207

Published on Saturday, 6th September 2008, 02:00 pm; Solved by 2719

For some positive integers k, there exists an integer partition of the form 4t = 2t + k,
where 4t, 2t, and k are all positive integers and t is a real number.

The first two such partitions are 41 = 21 + 2 and 41.5849625... = 21.5849625... + 6.

Partitions where t is also an integer are called perfect.
For any m 1 let P(m) be the proportion of such partitions that are perfect with k m.
Thus P(6) = 1/2.

In the following table are listed some values of P(m)

P(5) = 1/1
P(10) = 1/2
P(15) = 2/3
P(20) = 1/2
P(25) = 1/2
P(30) = 2/5
...
P(180) = 1/4
P(185) = 3/13

Find the smallest m for which P(m) 1/12345