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Power Equality Poser (Posted on 2015-03-29) Difficulty: 3 of 5
Find all nonzero integer solutions of this system of equations:

AA+B= B12 and, BA+B = A3

Prove that there are no others.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution Analytical Solution | Comment 1 of 4
1) Taking logs of both equations gives

   (A+B)*lnA = 12*ln(B)
   3*ln(A) = (A+B)*lnB
2) Dividing LHS by LHS and RHS by RHS is legitimate unless ln(A) - ln(b) = 0, and checking we see that (1,1) is in fact a solution.  Otherwise,

   (A+B)/3 = 12/(A+B)

3) Rearranging and solving gives A+B = +/- 6
   Clearly, A+B cannot equal -6, so A+B = 6
4) Substituting (A+B) = 6 in either equation and taking roots gives 
   A = +/- B^2
5) If A and B are both positive, the only solution is {4,2)
   If B is negative, the only solution is (9, -3)
   And there can be no solution where A is negative, because that makes abs(A) less than B, so A cannot = -B^2

6) (0,0) is not a solution because 0^0 is undefined.
    (-1,-1) satisfies the first equation but not the second.
 Only solutions are (1,1) and (4,2) and (9,-3)  

Edited on March 29, 2015, 10:11 am
  Posted by Steve Herman on 2015-03-29 09:56:54

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