Hello, I am Mahima. I would like to ask the following clarifications. If any one answered, I am so thankful to you. 

In which bases is 1111 a square?
b^3 + b^2 + b + 1 = n^2.
(b + 1)(b^2 + 1) = n^2.
We look at the gcd(b+1, b^2 +1) using the Euclidean algorithm.
And find that gcd(b+1, b^2 +1) = 2 if b is odd, but 1 if b is even.
If b is even, we have both (b + 1) and (b^2 + 1) a square.
But that is not possible as no positive squares differ by 1.
So b is odd, and both b + 1 and b^2 + 1 are even, so they are both twice a
square.
So we have:
b + 1 = 2a^2  and b^2 + 1 = 2c^2.

These are simultaneous diophantine equations.
We solve the one with least solutions and test these with the easier one.
The second is a Pellian equation.
The smallest solution is b = 7, c = 5.  This also satisfies the first.
So we have one solution.
base 7  1111 = 1 + 7 + 7^2 + 7^3 = 20^2.
Using the method for solving the Pellian, I can't find another solution
for both equations.  I may be able to produce a proof by induction that 
the solution is unique.
I have had a look at base 12, and think it might be a limited base. 
Please see what you can do there.

I want generalizations also.
Thanks in advance.
with LOVE,
Mahima.