Each of A, B, C and D is a nonzero integer such that:

A

^{2} + B

^{2} + C

^{2} = D

^{4}, and:

A + B + C = D

^{2}
Find the four smallest values of abs(A) + abs(B) + abs(C)

__Note__: abs(x) is the

** Absolute Value Function**
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**