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.