Is there a full square in the form XX?

Since such a number is X*(10^n + 1), let's examine the number 1000...01. If this is squarefree, then for our number to be a perfect square we would need X>=10^n+1, which won't work, so we seek an n to make 10^n+1 "squareful". The first such is n=11, where

10^11+1 = 11^2 * 23 * 4093 * 8779

so first have a look at the number 23*4093*8779, which will multiply 10^11+1 to give a square. This number turns out to be too small (only 9 digits), so we'd have extra 0's padding the number. So multiply by an appropriate square, say for instance 2^6, to get

X=52,892,561,984

Then 5,289,256,198,452,892,561,984 = 2^6 11^2 23^2 4093^2 8779^2 is one of the desired numbers.

As long as there are numbers of the form 10^n+1 which are not squarefree (I would expect there are infinitely many of them; indeed, it may be the case that they are only squarefree for finitely many n; but now I'm treading beyond my number theory knowledge), we can find more examples. Indeed, for each such n, we should be able to come up with at least several examples.

Ben's answer is great, just some comments:

There are an infinity of them and the smallest one is 1322314049613223140496=36363636364^2

A method to generate some of them is n^2 where

n=[10^(11k)+1]/11*10^s, for any k odd, where s is a shifting power.

This is because 10^(11k)+1=0(mod 11^2) and

n^2=[10^(11k)+1]/11^2*[10^(11k)+1]

*10^(2s)

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