All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Simultaneous Settlement IV (Posted on 2014-03-22) Difficulty: 3 of 5
Determine all possible ordered triplets (A,B,C) of integers satisfying this simultaneous equation:

A3 + B3 + C3 = 3 and:
A + B + C = 3

Prove that there are no others.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts computer exploration | Comment 1 of 2

The following program was manually stopped at a point where the total of A's and B's absolute values happened to reach 53,974:

   10   for T=2 to 99999999
   20     for A=1 to T\2
   30        B=T-A
   40        for Co1=-1 to 1 step 2
   50          for Co2=-1 to 1 step 2
   60            C=3-Co1*A-Co2*B
   70            if Co1*A*A*A+Co2*B*B*B+C*C*C=3 then print Co1*A;Co2*B;C
   80          next
   90        next
  100     next
  110   next
OK
run
 1  1  1
 4  4 -5
 4 -5  4
Break in 80
?t
 53974
OK

so, together with (-5, 4, 4), the ordered triplets are (1,1,1), (4,4,-5), and (4,-5,4).

Any further solutions would need |A| + |B| to exceed 53973, or actually any such pair from the three, due to the symmetry of the situation.


  Posted by Charlie on 2014-03-22 13:17:21
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information