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

Home > Just Math
A Cubic And Biquadratic Problem (Posted on 2006-06-18) Difficulty: 4 of 5
Find the three smallest positive integers (with the second number exceeding the first) such that the sum of the squares of the first two numbers is equal to the cube of the third number while the sum of the cubes of the first two numbers is equal to nine times the fourth power of the third number.

See The Solution Submitted by K Sengupta    
Rating: 3.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution computer solution (spoiler) Comment 1 of 1

The below program was halted at a total of 24576 for the first two numbers, so the shown solution (625, 1250, 125) and pseudosolution (1458, 1458, 162, where B does not exceed A) are the only two that exist for the total of the first two to be less than 24,576.

list
   10   for T=1 to 1000000
   20   for A=1 to int(T/2)
   30    B=T-A
   40    Cc=A*A+B*B
   50    C=int(Cc^(1/3)+0.5)
   60    if C*C*C=Cc then
   70      :if A*A*A+B*B*B=9*Cc*C then
   80      :print A,B,C
   90   next A
  100   next T
OK
run
 625     1250    125
 1458    1458    162
Break in 50
asave "cubbiq"
OK
cont
Break in 50
?t
 7965
OK
cont
Break in 50
?t
 12650
OK
cont
Break in 50
?t
 24576
OK

  Posted by Charlie on 2006-06-18 13:22:25
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 (9)
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