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 $b+c$?

RektTheNoob
Aug 8, 2017

#1**+1 **

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

CPhill
Aug 8, 2017