Which powers remain as powers when the digits are written in reverse?

Which non-trivial powers remain as non-trivial powers, if the digits are written in reverse ?

A power $a^b$ is called non-trivial, if $a,b>1$ holds.

Numbers ending with a $0$ are excluded to avoid leading zeros in the reverse number.

If we also exclude perfect squares , it seems that only palindromes are possible. The numbers upto $10^7$ are :

[$\ 8 , 343 , 1331 , 14641 , 1030301 , 1367631\$]

All the numbers are cubes with the exception $14641$ , which is a $4-th$ power.

• Question : Does a non-palindromic power of this kind exist ?

The non-palindromic squares upto $10^7$, for which the reverse is also a square, are :

144 441    2   2
169 961    2   2
441 144    2   2
961 169    2   2
1089 9801    2   2
9801 1089    2   2
10404 40401    2   2
10609 90601    2   2
12544 44521    2   2
12769 96721    2   2
14884 48841    2   2
40401 10404    2   2
44521 12544    2   2
48841 14884    2   2
90601 10609    2   2
96721 12769    2   2
1004004 4004001    2   2
1006009 9006001    2   2
1022121 1212201    2   2
1024144 4414201    2   2
1026169 9616201    2   2
1042441 1442401    2   2
1044484 4844401    2   2
1062961 1692601    2   2
1212201 1022121    2   2
1214404 4044121    2   2
1216609 9066121    2   2
1236544 4456321    2   2
1238769 9678321    2   2
1256641 1466521    2   2
1258884 4888521    2   2
1442401 1042441    2   2
1444804 4084441    2   2
1466521 1256641    2   2
1468944 4498641    2   2
1692601 1062961    2   2
4004001 1004004    2   2
4044121 1214404    2   2
4048144 4418404    2   2
4084441 1444804    2   2
4088484 4848804    2   2
4414201 1024144    2   2
4418404 4048144    2   2
4456321 1236544    2   2
4498641 1468944    2   2
4844401 1044484    2   2
4848804 4088484    2   2
4888521 1258884    2   2
9006001 1006009    2   2
9066121 1216609    2   2
9616201 1026169    2   2
9678321 1238769    2   2
?


Question : Do finite many or infinite many such non-palindromic squares exist ?

• The examples 144, 10404, 1004004, 100040004, . . . and 169, 10609, 1006009, 100060009, . . . give two infinite families. The squares are dense enough that one might also expect infinitely many "random" examples but that's probably not going to be proved any time soon. May 23 '16 at 13:16
• OK, this question has apparantly be answered. May 23 '16 at 13:17

The first pair of each length in the given list has the fom $$(10^k + 2)^2 = 10^{2k} + 4 \cdot 10^k + 4, \qquad (2 \cdot 10^k + 1)^2 = 4 \cdot 10^{2k} + 4 \cdot 10^k + 1,$$ and it's easy to see that these expressions comprise a pair of "nonpalindromic squares" for all $k \geq 1$.

• This was already mentioned by Noam D. Elkies May 23 '16 at 13:21
• @Peter: To be fair to Travis, the difference in posting time was only $2\frac12$ minutes. May 23 '16 at 13:24
• @Peter Yes, Noam posted his comment while I was writing my answer. May 23 '16 at 16:51