find infinitely many (or all) positive integers n so that n and rev(n) are perfect squares

Question

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?

Answer 1

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.

Consider the following construction :

$$ \begin{align}1{\underbrace {0\dots0}_{n \thinspace \text{digits}}2}^2&=1{\underbrace {0\dots0}_{n \thinspace \text{digits}}4{\underbrace {0\dots0}_{n \thinspace \text{digits}}4}}\\\\ 2{\underbrace {0\dots0}_{n \thinspace \text{digits}}1}^2&=4{\underbrace {0\dots0}_{n \thinspace \text{digits}}4\underbrace {0\dots 0}_{n \thinspace \text{digits}}1}\end{align} $$

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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?

Possible palindromic number family :

$$ \begin{align}1{\underbrace {0\dots0}_{n \thinspace \text{digits}}1{\underbrace {0\dots 0}_{n \thinspace \text{digits}}}1}^2&=1{\underbrace {0\dots 0}_{n \thinspace \text{digits}}2{\underbrace {0\dots 0}_{n \thinspace \text{digits}}3}}{\underbrace {0\dots 0}_{n \thinspace \text{digits}}2{\underbrace {0\dots 0}_{n \thinspace \text{digits}}1 }}\end{align} $$

Answer 2

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 $\sum_{i}A_{i}A_{K-i} \ge 10$.

Answer 3

You ended on the questions how can one find all perfect squares that are palindromes or infinitely many such squares?
and were given two ways to generate the latter, so I'll address the former.

The complete list of numbers up to a specific upper bound is currently only heuristically found.
A Mathematica procedure to obtain all such numbers $\leq n^2$ is

Select[Range[0, n]^2, IntegerQ[Sqrt[IntegerReverse[#]]]&]\

A procedure to obtain their square roots $\leq n$ is

Select[Range[n], IntegerQ[Sqrt[FromDigits[Reverse[IntegerDigits[ #^2]]]]] &]

Below is the list of all perfect squares $\leq 1'000'000$ whose reverse is also a perfect square, with palindromic numbers in $\color{blue}{blue}$, non-palindromic in $\color{red}{red}$, and multiples of $10$ in black, as they yield reverses with leading zeroes.

Notably $\color{blue}{26^2=676}$, $\color{red}{99^2=9801}$, $260^2=67600$, $\color{blue}{264^2=69696}$, $\color{blue}{307^2=94249}$, $330^2=108900$ and $\color{blue}{836^2=698896}$ stand out as numbers that are not found with the generalization mentioned in previous answers.

roots	perfect squares	rev(perfect squares)
A102859	A061457
$\color{blue}{0}$	$\color{blue}{0}$	$\color{blue}{0}$
$\color{blue}{1}$	$\color{blue}{1}$	$\color{blue}{1}$
$\color{blue}{2}$	$\color{blue}{4}$	$\color{blue}{4}$
$\color{blue}{3}$	$\color{blue}{9}$	$\color{blue}{9}$
10	100	001
$\color{blue}{11}$	$\color{blue}{121}$	$\color{blue}{121}$
$\color{red}{12}$	$\color{red}{144}$	$\color{red}{441}$
$\color{red}{13}$	$\color{red}{169}$	$\color{red}{961}$
20	400	004
$\color{red}{21}$	$\color{red}{441}$	$\color{red}{144}$
$\color{blue}{22}$	$\color{blue}{484}$	$\color{blue}{484}$
$\color{blue}{26}$	$\color{blue}{676}$	$\color{blue}{676}$
30	900	009
$\color{red}{31}$	$\color{red}{961}$	$\color{red}{169}$
$\color{red}{33}$	$\color{red}{1089}$	$\color{red}{9801}$
$\color{red}{99}$	$\color{red}{9801}$	$\color{red}{1089}$
100	10000	00001
$\color{blue}{101}$	$\color{blue}{10201}$	$\color{blue}{10201}$
$\color{red}{102}$	$\color{red}{10404}$	$\color{red}{40401}$
$\color{red}{103}$	$\color{red}{10609}$	$\color{red}{90601}$
110	12100	00121
$\color{blue}{111}$	$\color{blue}{12321}$	$\color{blue}{12321}$
$\color{red}{112}$	$\color{red}{12544}$	$\color{red}{44521}$
$\color{red}{113}$	$\color{red}{12769}$	$\color{red}{96721}$
120	14400	00441
$\color{blue}{121}$	$\color{blue}{14641}$	$\color{blue}{14641}$
$\color{red}{122}$	$\color{red}{14884}$	$\color{red}{48841}$
130	16900	00961
200	40000	00004
$\color{red}{201}$	$\color{red}{40401}$	$\color{red}{10404}$
$\color{blue}{202}$	$\color{blue}{40804}$	$\color{blue}{40804}$
210	44100	00144
$\color{red}{211}$	$\color{red}{44521}$	$\color{red}{12544}$
$\color{blue}{212}$	$\color{blue}{44944}$	$\color{blue}{44944}$
220	48400	00484
$\color{red}{221}$	$\color{red}{48841}$	$\color{red}{14884}$
260	67600	00676
$\color{blue}{264}$	$\color{blue}{69696}$	$\color{blue}{69696}$
300	90000	00009
$\color{red}{301}$	$\color{red}{90601}$	$\color{red}{10609}$
$\color{blue}{307}$	$\color{blue}{94249}$	$\color{blue}{94249}$
310	96100	00169
$\color{red}{311}$	$\color{red}{96721}$	$\color{red}{12769}$
330	108900	009801
$\color{blue}{836}$	$\color{blue}{698896}$	$\color{blue}{698896}$
990	980100	001089
1000	1000000	0000001

Answer 4

If n and rev(n) are perfect squares, then 100n is a perfect square, rev(100n) = rev(n) is a square, so 100n is also a solution. On the other hand, if 100n is a perfect square, then so is n, and rev(100n) = rev(n) is als a square. So we can ignore trivial solution ending in an even number of 0’s, and numbers ending in an odd number of 0’s are not square. We only look for n not ending in 0.

For an exhaustive search to find all solutions n <= N for some large fixed N, we iterate through all squares <= N that are not multiples of 10. Now every square ends in 1, 4, 5, 6 or 9, so we can ignore all squares where rev(n) ends in 2, 3, 7 or 8, which is all squared starting with 2, 3, 5, 7 or 8.

We can do the same for the last 2, 3, 4 or 5 digits and reduce the number of squares to be checked massively.