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Quadratic and Quartic Query (Posted on 2015-09-06) Difficulty: 3 of 5
Each of A, B, C and D is a nonzero integer such that:

A2 + B2 + C2 = D4, and:
A + B + C = D2

Find the four smallest values of abs(A) + abs(B) + abs(C)

Note: abs(x) is the Absolute Value Function

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution Analytical Approach (spoiler?) | Comment 1 of 3
Squaring the 2nd equation and subtracting the first gives

2AB + 2AC + 2BC = 0

They cannot all be positive or all be negative, so without loss of generality, assume A and B are positive and C is negative.

Solving for C gives
    C = -AB(A+B).
    
If A = 1, then C = -B/(B+1) which has no solution for B positive

If A = 2, then C = -2B/(B+2) which leads to (2,2,-1).  But A+B+C = 3, which is not square

If A = 3, then C = -3B/(B+3) which leads to (3,6,-2).  But A+B+C = 7, which is not square

If A = 4, then C = -4B/(B+4) which leads to (4,4,-2).  But A+B+C = 6, which is not square
                          it also leads to (4,12,-3).  But A+B+C =13, which is not square
                          
      Too bad the problem wasn't A^2+B^2+C^2 = D^2 and A+B+C = D
      
It is obvious that B must be a multiple of A, so let B = Ak
Then C = -A(Ak)/(Ak + A) = -Ak/(k+1).  k+1 must divide A.

If A = 5, then k can only be 4.  This leads to (5,20,-4).  A+B+C = 29

If A = 6, then k can only be 1,2,or 5.  
     k = 1 leads to (6,6,-3).  A+B+C = 9!!! First solution  |A|+|B|+|C| = 15
     k = 2 leads to (6,12,-4).  A+B+C = 14. Not a square 
     k = 5 leads to (6,30,-5).  A+B+C = 31. Not a square

But wait!
  (6,6,-3) is just 3 * my A=2 case  A+B+C was not a square,adding to 3, but 3*(A+B+C) is a square.
  
More solutions include
  7 * (3,6,-2) = (21,42,-14).  |A|+|B|+|C| = 77
  6 * (4,4,-2) = (24,24,-12).  |A|+|B|+|C| = 60
 13 * (4,12,-3)= (52,156,-39). |A|+|B|+|C| = 247
 
Are these the 4 lowest |A|+|B|+|C|?
I think not, because I can get another solution based on my first solution:

 9 * (6,6,-3) = (54,54,-27).  |A|+|B|+|C| = 135
 
There might be lower solutions that I haven't hit yet, but I will let the computers look. 

I propose an answer of 15, 60, 77 and 135.
  


Edited on September 6, 2015, 1:24 pm
  Posted by Steve Herman on 2015-09-06 12:21:09

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