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i wanna formulate some intresting results as other great mathematicians did

so i observed a sequence of perfect squares which are palindrome

i found in one digit we have $1,4,9$

in two digit no such number exist as

number would be divisible by $11$hence by$121$

in $3$ digit

$$121$$$$484$$$$676$$

but i don't get any sign of limited terms so i wanna help to prove or disprove that there exist infinite terms

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    http://www.fengyuan.com/palindrome.html – 2017-01-08
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    The numbers $121,10201,1002001,100020001,\cdots$ are all squares. – 2017-01-08
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    A palindromic number with even length (i.e. the number of digits is even) is [a multiple of 11](https://math.stackexchange.com/questions/697096/). See also https://oeis.org/A002779 – 2017-01-08
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    "This paper originated from the observation that although there are infinitely many decimal numbers whose squares, cubes, and fourth powers are palindromes" ; "10000..0001 (with $i$ zeroes) is obviously such a numerb for every $i \geq 1$", in [G. J. Simmons, On palindromic squares of non-palindromic numbers, J. Rec. Math., 5 (No. 1, 1972)](https://oeis.org/A002778/a002778.pdf). – 2017-01-08

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