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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 ?

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    $\begingroup$ 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. $\endgroup$ May 23 '16 at 13:16
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    $\begingroup$ OK, this question has apparantly be answered. $\endgroup$
    – Peter
    May 23 '16 at 13:17
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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 $k \geq 1$.

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

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