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 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 $k \geq 1$.