Suppose $b$ and $c$ are positive integers. When $b^2$ is written in base $c$, the result is $121_c$. When $c^2$ is written in base $b$, the result is $71_b$. What is $b+c$?
We have that
c^2 + 2c + 1 = b^2 (1)
and
7b + 1 = c^2 (2)
Factoring the first, we have that
(c + 1)^2 = b^2 and since b is positive
c + 1 = b sub this into (2)
7(c + 1) + 1 = c^2 simplify
7c + 8 = c^2 rearrange
c^2 - 7c - 8 = 0 factor
(c - 8)(c + 1) = 0
Setting both factors to 0 and solving for c, we have that c = 8 or c = -1
But c is positive so c = 8 and b = c + 1 = 9
Proof
c^2 + 2c + 1 = 64 + 16 + 1 = 81 = b^2
And
7b + 1 = 64 = c^2
So b + c = 17