For a positive integer n, let rev(n) denote the integer obtained by reversing the digits of n. Find infinitely many (or all) positive integers n so that n and rev(n) are perfect squares.
The problem is similar in nature to some contest problem, though I'm not sure if there are neat tricks for solving it.
Call a positive integer n satisfying the problem's condition good. Note that all 3 1-digit perfect squares n satisfy the condition. Also, by inspection, no two digit numbers satisfy the condition. If $100a+10b+c$ and $100c+10b+a$ are both perfect squares, then $99(c-a)$ is the difference of the perfect squares. Note that $144$ is good. We can check to see which multiples of 99 from 0 to $99\cdot 9$ are the difference of two squares. Alternatively, we can manually check all 3-digit perfect squares to find the good ones. But how can I generalize my approach to numbers with arbitrarily many digits? All perfect squares that are palindromes are obviously good, but how can one find all perfect squares that are palindromes or infinitely many such squares?
Generalizing a previous answer, the square of any sum of the form $\sum_{i=0}^{n} A_i 10^i$ works as long as (1) the $A_i$ are all $0$, $1$, $2$, or $3$; (2) the nonzero $A_i$ are sufficiently far apart and irregularly spaced (*); and (3) there's no pair $i\neq j$ such that $A_i A_j > 5$. (So if there's a $3$, there can only be one; and there can't be both a $3$ and a $2$.) For instance, $$ 1001000200002^2=1002001400404044000800004=rev(2000020001001^2). $$ Note here that the squaring operation and the reversing operation commute (i.e., the reversed square of $x$ is the square of the reverse of $x$). Are there examples where that isn't the case?
(*) More precisely, in the family of solutions described here, the squaring operation doesn't produce carries: there is no $K$ such that $\sum_{i}A_{i}A_{K-i} \ge 10$.