Four digit palindromes

Let the palindromes be $\overline{abba} = 1001a + 110b$ , where $a$ and $b$ are single-digit integer. Since $1001a+110b$ is divisible by $11$ , the powers we look for must be of the form $(11n)^p$ , where $n$ and $p$ are positive integers. For $p=2$ , we note that none of the $121n^2$ is a four-digit palindrome. For $p=3$ , we have $1331n^3$ and note that when $n=1$ we get $1331$ as a four-digit palindrome, For $n > 1$ the cube number has five or more digits. For $p>3$ , the power number is larger than five-figure. Therefore $1331$ is the only perfect power and the answer is $\boxed 1$ .

Let the four digit palindromic number be $\overline {ABBA}=1001A+110B=11(91A+10B)$ . Then $91A+10B\leq 909$ . Also $91A+10B$ must either be of the form $11a^2, a\leq 9$ , or $11^3=1331$ . Since $22^3=10648>9999$ , no other perfect powers are possible. A direct check shows that $\dfrac{91A+10B}{11}$ is not a perfect square for any integer value of $0\leq A, B\leq 9$ . So the only possible number is $\boxed {1331=11^3}$ which is four digit palindromic. Hence the answer is $\boxed 1$ .