Pete Inglesby from Wolverhampton; Patrick Snow, age 16, Dame Alice Owens School, Potters Bar; Yatir Halevi, age 18, Maccabim-Reut High-School, Israel; Andrei Lazanu, age 12, School 205, Bucharest, Romania all solved this problem. Here is Pete's solution:
Consider numbers of the form How many such numbers are perfect squares?
| b | b 2 | Last digit of b 2 |
| 0 | 0 | 0 |
| 1 | 1 | 1 |
| 2 | 4 | 4 |
| 3 | 9 | 9 |
| 4 | 16 | 6 |
| 5 | 25 | 5 |
| 6 | 36 | 6 |
| 7 | 49 | 9 |
| 8 | 64 | 4 |
| 9 | 81 | 1 |
This table shows that the squares of all integers between 0 and 9, and therefore all integers, end in either 1, 4, 5, 6 or 9.
No perfect squares end in 3. Therefore, u(n) with n > 4 can never be a perfect square as it has been shown to always end in 3.
All that remains is to find u(n) for all integers of 4 or less:
Only u(1) and u(3) are perfect squares, and so they are the only sums of factorials to be perfect squares.