Is a 2n-digit number with identical halves a square number?
Note that such number is c(10^n+1) where 10^(n-1) <= c < 10^n.
Let 10^n+1 = m*t^2, where t^2 is the largest square dividing 10^n+1. Then c >= m.
If t = 1 (10^n+1 is squarefree) then c >= m = 10^n+1, so no such number is square.
Otherwise let c = m*b^2, for any b with t/sqrt(10) < b < t.
Then c(10^n+1) = (m*b^2)(m*t^2) = (mbt)^2
and 10^(n-1) <= c < 10^n.
In other words, there exists a 2n-digit square with identical halves if and only if 10^n+1 is non-squarefree.
Example: n = 11 (the least n such that 10^n+1 is not squarefree)
10^11+1 = 100000000001 = (11^2)*23*4093*8779 = 826446281*11^2
So m = 826446281 and t = 11.
t/sqrt(10)~3.47 < 4, so let b=4. (b can be anything between 4 and 10 inclusive, for 7 solutions)
c = m*b^2 = 13223140496.
mbt = 36363636364
Therefore 36363636364^2 = 13223140496_13223140496.
Note that mbt + m(t-b)t = m*t^2 = 10^n+1.
Hence the 10^n+1_complement of a solution's base is the base of another solution (if t-b is greater than t/sqrt(10)).
e.g. 63636363637^2 = 40495867769_40495867769 (t-b = 11-4 = 7)
The next n with a solution is n = 21 (t = 7) with 4 solutions (3 <= b <= 6).
b = 3: 4285714285714285714292 = 183673469387755102041_183673469387755102041.
Other n with solutions: n = 33 (t = 11), n = 39 (t = 13), n = 55 (t = 11), n = 63 (t = 7)
In general, there are families of solutions: n = 22k+11 (11|t), n = 42k+21 (7|t), n = 78k+39 (13|t), and so on.
Are there any 2n-digit cubes or higher powers with identical halves? Such number requires that 10^n+1 has at most
a squarefree part of sqrt(10^n-1). However, the squarefree part of 10^n+1 seems to be very close to 10^n+1 for n>1,
if not 10^n+1 itself.