If you consider the square numbers in base 10, very few of them end in repeated digits, apart from the trivial case of all squares of multiples of 10 ending in multiple 0’s.

Square numbers which end in 1 or 9 always have an even number in the 10s column. (01, 09, 49, 81, 121, 169…) Square number which end in 6 always have an odd number in the 10s column. (16, 36, 196, 256…) Square numbers which end in 5 always in fact end with 25, so never with 55.

Square numbers which end in 4, however, always have an even number in the 10s column (04, 64) and so we get plenty ending in ..-44 (starting with 144 which is 12 squared) and even …444 (starting with 1444 which is 38 squared). But there it stops. If the last three digits are 4’s, the number in the thousands column is odd: 462 squared is 213444, 538 squared is 289444, etc.

What about other bases? Well, in binary, every even number’s square ends with at least two zeros (because they are all multiples of four), and every odd number’s square ends in …001 (because they are always one more than a multiple of eight).

In hexadecimal we start off promisingly:

Now we have to go up considerably in scale, and there aren’t any squares ending in more than five repeated 4’s, as the sixth last digit is always odd (1’s and 9’s, however, continue)

And I guess that we can keep going with steadily increasing numbers of 1’s and 9’s (this is probably trivial enough to prove).

Counting in an odd-numbered base, it is also pretty easy to build up repeated digits as the distinction between odd and even numbers no longer applies.