# Perfect Powers

How many $3$ digit numbers $N$ can be expressed as $\overline{aba} = p^q$, where $p$ and $q$ are positive integers and $q>1$

$A$ $request$ : Please explain your steps a bit if you use Modular Arithmetic because I ain't much used to it.

Note by Abhimanyu Swami
6 years, 12 months ago

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Again I believe there is no better choice than a case-by-case analysis. I rephrase the problem to find the number of 3-digit palindrome perfect power. First note that no power of 2 is a 3-digit palindrome. Since $3^7>1000$, we just need to check until $q=6$. Furthermore, if $q$ is even, $p^q$ is a perfect square and we may assume $q=2$. Now we consider $q=5,3,2$.

Case 1: $q=5$

$4^5>1000$, and $3^5$is not a palindrome.

Case 2: $q=3$

Just manually check for $1\leq p\leq 9$, and we find only $7^3=343$ is a palindrome.

Case 3: $q=2$

Recall that the quadratic residues modulo $10$ are $1,4,9,6,5$. Substitute $a=1,4,9,6,5$ and check to find a $b$ such that $\overline{aba}$ is a square. One arrives at $11^2=121, 22^2=484, 26^2=676$.

Therefore we conclude there are 4 3-digit palindrome perfect powers.

- 6 years, 12 months ago

Thanks again.

- 6 years, 11 months ago

You're welcome.

- 6 years, 11 months ago

The answer is 4. (7^3 = 343, 11^2 = 121, 22^2 = 484, 26^2 = 676) *Brute force. (I'll try to prove this later.)

- 6 years, 12 months ago