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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
avatar+90088 
+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

 

 

cool cool cool

CPhill  Aug 8, 2017

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