I did this problem today after school!!! xD Well, if that number exists, first, it doesnt have the digits 2,3,7,8 (because no square can finish in that numbers, and some permutations take numbers with final digit 2,3,7,8). So that number has to has this digits: 0,1,4,5,6,9. But if it has a 0, then we can take any digit (for example 9) and take a permutation finishing in 90. This number is divisible by 10, but no by 100 so it cant be a square. Now the number has to has this digits: 1,4,5,6,9 But if it has a 5, we can take some digit (obviously not equal to 2, for example 7) and we take the permutation finished in 75. This number is divisible by 5, so if its a perfect square it must finish in 25, contradiction. So the number has only the digits 1,4,6,9 Using the same idea as before, take a permutation finishing in 14. We know that the number squared to get that permutation has to be even, so its (10a+2b)^2 (a natural and $ 0\le b\le 4$) and $ (10a+2b)^2\equiv 40ab+4b^2$ its a number with even number in the tens... so 14 and 94 fails. Then the number can only have the digits 6 or 4. If the number has both numbers, then exists a permutation that finishes in $ 46\equiv 6(8)$ but 6 is a non-cuadratic residue modulo 8, contradiction. So the number must have only 4's or only 6's But a number only consisting of 444..444 cant be a perfect square because of modulo 16 ($ 444...44\equiv 4444\equiv 12(16)$) but 12 is a non-cuadratic residue modulo 16 so it must consist of only 6's. But $ 666...666\equiv 2(8)$ and 2 is a noncuadratic residue modulo 8 so the initial number cant exist.

There are obvious flaws. All that is proved at some point is that $1$ and $4$ cannot appear both as digits, and $9$ and $4$ cannot appear both as digits; this does not rule out the digits $1$ and $9$.

Here's my solution.

Attachments:
JBMO_2006.docx (16kb)