Square numbers are defined and calculated by this extraordinary intricate and excruciatingly complex formula.
So, this line is for experts only _{}
To the right a graphical representation of the first three square numbers to whet the beginner's appetite.
( base ) x ( base ) or base ^{2}
square 1  square 4  square 9 
PLAIN TEXT SQUARES  PLAIN TEXT SSP 
This basenumber 982.503.990.036.767.976.718.486.749.830.913 has 33 digits yielding the following palindromic square record number with even number of digits 965.314.090.438.169.467.661.632.561.053.978.879.350.165.236.166.764.961.834.090.413.569 having a staggering length of 66 digits. 
This world record was achieved using CUDA code written by Robert Xiao and no longer on Rust. Recently he generalized the program to handle arbitrary quadratics. CUDA is a programming language, or more properly a programming toolkit, for writing software to run on GPUs rather than CPUs. It runs about 50 times faster on our GPUs though the logic of the code follows the Rust version closely. I asked Robert now that his CUDA is running at warp speed how far it would reach. He answered that as for 70 digits the time estimate on that is around ~400 days on one of our GPUs. 60 digits is about two days of GPU time, and it’ll go up by a factor of 10 every 4 digits. Doable but it’ll be a pretty decent power bill :) “Maybe we could get some palindrome enthusiasts together”, as David Griffeath put it, “and get a distributed computation going.” The program is very amenable to divideandconquer approaches. 
Case Square  Change of variables  CUDApalin parameters  Base Correction 
base ver 1  n.a.  A B C → 1 0 0  base = CUDAbase 
base ver 2  n = m + 1  A B C → 1 2 1  base = CUDAbase + 1 
Mike Keith made an original study of the palindromic squares.
He was able to classify and enumerate them in a logical manner
and went even further as he tried also to create a general formula for
calculating the number of certain classes of palindromic squares,
except for the 'sporadic' solutions. Wonderful!
[Please refer to the “JRM” source detailed in section 'Sources Revealed'].
Mike had most trouble with the Asymmetric Root Family but thanks to
Dave Wilson's “lemma” he made significant progress.
What are the conditions for creating a pseudopalindrome with a palindromic square ? Open this window and read David W. Wilson's answer. 
Unlike Palindromic Triangulars where it is impossible to
predict a next higher one, whether its basenumber is palindromic or not,
with the Palindromic Squares (and Cubes) we have an opposite situation.
Finding a next higher number is very easy.
Start e.g. with the number 11. Then repeatedly add a zero between the two 'ones' and square them.
A pattern emerges that can go on forever.
Base  Square 

11  121 
101  10201 
1001  1002001 
10001  100020001 
There are other numbers with the same properties (e.g. 10101). The repunit numbers like '111' looks also a good candidate
for finding palindromic squares (not for cubes though) but the expansion doesn't go on forever. Nine 'ones' still produce
a palindrome when squared, but not ten 'ones'. It's a near miss as only the number 8 doesn't show up in the left part of the square.
This anomaly stays with longer repunits as well, so we must abandon this candidate.
If we take resort to higher basenumbers only then we can extend the pattern (for a while)!
Base  Square 

1  1 
11  121 
111  12321 
1111  1234321 
11111  123454321 
111111  12345654321 
1111111  1234567654321 
11111111  123456787654321 
111111111  12345678987654321 
1111111111  1234567900987654321 = NOT palindromic in base 10 ! 
Here is how Natalie Dickendasher [ November 4, 2005 ] (email) figures them out quickly in her head and impresses friends or boss :
The highest digit of the sequence will always be the quantity of 1's in the number your are squaring.
As an example lets start with my salary for next year (yah right): what is 111111 squared ?
EASY !! This number has a total of 6 digits. Therefore the highest number in the palindrome will be 6
so count to 6 and then back down. The answer is 12345654321 [35].
Finding large palindromic squares is easy if you start with some palindromic basenumbers as explained above.
So, let's concentrate on the difficult ones namely those with a non palindromic basenumber. The first six are
(26 [7]) (264 [13]) (307 [14]) (836 [15]) (2285 [19]) and (2636 [20]) and can be found without too much trouble with a spreadsheet or calculator.
Mike Bennett made an attempt in finding out what his chances are in discovering palindromic squares and palindromic cubes. Here is his mathematical analysis and the probability of finding those interesting palindromic polygonal numbers. 
Square Palindrome_(SP)_length  1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30  31  32  33  34  35  

A263618  # SP's  4  0  3  0  7  1  5  0  11  0  5  1  19  0  13  1  25  0  18  0  48  1  31  0  70  1  44  2  105  0  70  1  153  1  98 
# SSP's    0  1  0  2  1  2  0  3  0  0  1  5  0  4  1  1  0  2  0  6  1  4  0  4  1  1  2  3  0  5  1  5  1  4  
# odd_length SSP's    1  2  2  3  0  5  4  1  2  6  4  4  1  3  5  5  4  
# even_length SSP's  0  0  1  0  0  1  0  1  0  0  1  0  1  2  0  1  1  
Continued...  Square Palindrome_(SP)_length  36  37  38  39  40  41  42  43  44  45  46  47  48  49  50  51  52  53  54  55  56  57  58  59  60  61  62  63  64  65  66  67  68  69  70 
A263618  # SP's  3  209  0  132  0  291  1  181  1  384  0  234  2  496  1  301  1  636  0  383  0  798  1  474  1  981  0  577  0  1196  2         
# SSP's  3  2  0  1  0  4  1  4  1  4  0  1  2  2  1  1  1  5  0  4  0  2  1  3  1  6  0  1  0  3  2          
# odd_length SSP's  2  1  4  4  4  1  2  1  5  4  2  3  6  1  3      
# even_length SSP's  3  0  0  1  1  0  2  1  1  0  0  1  1  0  0  2     
Basenumber(BN)_length  1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30  31  32  33  34  

A263616  A263617  # BN's giving SP's  4  3  8  5  11  6  19  14  25  18  49  31  71  46  105  71  154  101  209  132  292  182  384  236  497  302  636  383  799  475  981  578  1201    
# BN's giving odd_length SP's    3  7  5  11  5  19  13  25  18  48  31  70  44  105  70  153  98  209  132  291  181  384  234  496  301  636  383  798  474  981  578  1199    
# BN's giving even_length (S)SP's    0  1  0  0  1  0  1  0  0  1  0  1  2  0  1  1  3  0  0  1  1  0  2  1  1  0  0  1  1  0  0  2    
# BN's giving sporadics (↑)+(↓)    1  3  2  3  1  5  5  1  2  7  4  5  3  3  6  6  7  2  1  5  5  4  3  3  2  5  4  3  4  6  1  5    
# BN's giving odd_length SSP's    1  2  2  3  0  5  4  1  2  6  4  4  1  3  5  5  4  2  1  4  4  4  1  2  1  5  4  2  3  6  1  3    
# BN's giving NSSP's    2  5  3  8  5  14  9  24  16  42  27  66  43  102  65  148  94  207  131  287  177  380  233  494  300  631  379  796  471  975  577  1196  698  
A263614  # BN's binaries → B    1  2  2  4  4  8  8  16  15  30  26  52  42  84  64  128  93  186  130  260  176  352  232  454  299  598  378  756  470  940  576  1152  697  
A142150  # BN's ternaries → T    0  1  0  2  0  3  0  4  0  5  0  6  0  7  0  8  0  9  0  10  0  11  0  12  0  13  0  14  0  15  0  16  0  
A007573  # BN's asymmetrics → A    0  0  0  0  0  1  0  2  0  5  0  6  0  9  0  10  0  10  0  15  0  15  0  16  0  18  0  24  0  18  0  26  0  
A000034  # BN's even_root → E    1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1 
Can we find formulas for all or some of these sequences ?

“I've been feeling remiss about not showing you how that messy formula for the number of binaries doesn't involve fancy math, but rather just a bunch of unpleasant, unenlightening algebra. So I've attached a handwritten sketch of the computations.
The first step, in (1) and (2) is to change variables and express the formulas for the even and odd cases 
Click here to view some entries to the table about palindromes. 
Squares  Acquaint yourself with SQUARES  by Richard Phillips 

Square Number  From Eric Weisstein's Math Encyclopedia 
Ask Dr. Math  Someone gave me different numbers to find the square root of, including 14641. My answer was 121. Something special about this is that both numbers are read the same left to right as right to left. Do you know any other numbers in which both the number and its square root are this way ? 
A very interesting webpage by William Rex Marshall (19302010) is titled “Palindromic Squares”
and gives an overview of the first 67 palindromic squares known up to May 28, 2001.
The classification of these squares as set out by Michael Keith in the JRM from 1990 is also revealed.
http://www.geocities.com/williamrexmarshall/math/palsq.html (archived)
The second highest one (3069306930693 [263]) is copied from Martin Gardner's book
“The Ambidextrous Universe”
see page 40.
Another source “Curious and Interesting Numbers” by David Wells, page 185, provided me the basenumber 798644 [37].
While surfing the Internet I came across the work of Keith Devlin namely
All the Math that's Fit to Print.
In chapters 17, 75 and 110 he printed some palindromic squares he got from other readers.
The highest one that originates from his publication is 6360832925898 [264].
Gustavus J. Simmons published an article about palindromic squares in the “Journal of Recreational Mathematics”,
J. Rec. Math., 3 (No. 2, 1970), 9398, titled “Palindromic Powers”.
OEIS Source (annotated scanned copy) → https://oeis.org/A002778/a002778_2.pdf
Gustavus J. Simmons published an article about palindromic squares in the “Journal of Recreational Mathematics”,
Volume 5, No. 1, 1972, pp. 1119, titled “On Palindromic Squares of NonPalindromic Numbers”.
OEIS Source (annotated scanned copy) → https://oeis.org/A002778/a002778.pdf
Mike Keith (email) (website) published an article about palindromic squares in the “Journal of Recreational Mathematics”,
Volume 22, Number 2  1990, pp. 124132, titled “Classification and enumeration of palindromic squares”,
and was followed by an article by Charles Ashbacher, pp. 133135, “More on palindromic squares”.
OEIS Source (annotated scanned copy) → https://oeis.org/A002778/a002778_1.pdf
I found Feng Yuan's record palindromic square of 55 digits [3992] in the following archived blog
http://blogs.msdn.com/fyuan/archive/2008/01/31/stsapalindromicwordiwillremember.aspx
Click here to display the full listing up to length 31. Click here to display the sporadics listing of all such square palindromes (SSP). Click here to display the subsets of palindromic squares. 
Careful observation of some numbers in the list lead me to the discovery of ever greater nonpalindromic basenumbers whose squares are palindromic.
Specially the numbers with indexes [48], [85], [90] and [141] attracted my attention
because of the CORE number 091 in them.
[48] 1109111
[85] 110091011
[90] 111091111
[141] 11000910011
A second promising CORE number exists. It is 09901 and it shows up from [131], [219], [229], ....
[131] 10109901101
[219] 1010099010101
[229] 1011099011101
[346] 100110990111001
[354] 101000990100101
[364] 101010990110101
[536] 10010109901101001
[553] 10100009901000101
[563] 10100109901100101
[570] 10101009901010101
I think a new third CORE number is emerging... It is 0999001
Watch out from
[340] 100109990011001
[531] 10010099900101001
[545] 10011099900111001
Quite interesting. Let's put these three CORE numbers in a row :
091
09901
0999001
David W. Wilson took up the thread where I left off.
Read what he posted to me about these core 'pseudopalindromes'
[ February 16, 1998 ].
If we allow a digit "n" with value "–1", then 091 is equal to the palindrome 1n1,
while 09901 = 10n01, 0999001 = 100n001, etc. Thus
10110n01101 x 10110n01101  10110n01101 10110n01101.. 10110n01101... n0nn010nn0n..... 10110n01101....... 10110n01101........ 10110n01101..........  102210100272001012201
And what about this repeat pattern of the item 3069
and that is finalized alternatively with a chopped off and rounded version of it namely a 3 or 307...
[4] 3Each time the length increases with 2
[14] 307
[31] 3069 3
[56] 3069 307
[95] 3069 3069 3
[161] 3069 3069 307
[263] 3069 3069 3069 3
3069 occurs as displacements to powers of ten so that this number forms the nearest (probable) prime to those axes :
10^{1001} – 3069 is the largest prime of 1001 digits
10^{1001} – 3069 10^{5068} – 3069 10^{7565} – 3069 10^{19217} – 3069 10^{28898} – 3069 
10^{1555} + 3069 10^{2399} + 3069 10^{5016} + 3069 10^{5063} + 3069 10^{9896} + 3069 
NEVER ODD OR EVEN Read title backwards please !
It looks as though the decent sufficientlyrandom nonpatternlike unpredictable nonpalindromic basenumbers become very rare.
In the table (see Full Listing or the Subsets) I highlighted them using a lightblue background in the sidecells.
And even then a further distinction can be made. Consider the even and odd length values for a minute.
Sporadic Palindromic Squares with an even length are just a plain minority. Until now I can only list twenty of them.
These are highlighted in the table by means of a aquamarine background color in the sidecells.
Score 'subset E' = 20
I 'almost' solved a very ancient problem namely : squaring the circle (pi = 3,1415....)
MORELAND PI welcomes PALINDROMES
[415] 314155324482867
This basenumber starts with the the number 2201 which is (to me) the only known nonpalindromic basenumber
of a palindromic cube ! The cube of 2201 equalling 10662526601.
[481] 2201019508986478
Base  Square 

1  1 
101  10201 
10101  102030201 
1010101  1020304030201 
101010101  10203040504030201 
10101010101  102030405060504030201 
1010101010101  1020304050607060504030201 
101010101010101  10203040506070807060504030201 
10101010101010101  102030405060708090807060504030201 (° see conjecture below) 
1010101010101010101  10203040506070809100908070605004030201 = QUITE UNpalindromic ! 
CONFIRMATION 

10101010101010101 is indeed the smallest integer that generates a palindromic square 102030405060708090807060504030201 that is also pandigital (contains all the ten digits). 
Here is a good place to talk about a famous conjecture that nobody
could prove or disprove for more than 80 years.
I must thank Mike Keith for letting me know this fact through an email message.
Mike has heard that this conjecture originates from
L. E. Dickson's famous “History of the Theory of Numbers”.
Can someone with access to this source confirm this ?
History :
My table lists exhaustively all palindromic squares up to lengths 31.
I also listed all the palindromic squares of length 32.
Palindromic squares of 'even' nature seem to be very popular among the record
breaking submitters. They told me there are no gaps for the 'even' length 32 !
So, in fact I had to finish checking only length 33...
Of which I checked all squares beginning with 1, 5 and 9 !
( and 50 % of those beginning with digits 4 and 6 ).
I collected a total of 151 palindromic squares beginning with digit 1.
(Un)lucky for me Dickson's number 102030405060708090807060504030201
( 10101010101010101^{2} ) [572] is the 86^{th} one. Anyway after checking these 86 squares
I could proudly announce that the conjecture was finally proven [ May 3, 1998 ].
The second smallest number [633] that answers to these rules is this one (also of length 33) :
11843191515764821^{2} = 140261185279083838380972581162041.
Alain Bex (email) gave also some palindromic squares in base 12.
Link to Palindromic Numbers other than Base 10
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