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

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