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

Home > Numbers
p^7 + q^3 = r^2 (Posted on 2011-01-09) Difficulty: 3 of 5
Determine all possible triplets (p, q, r) of positive integers that satisfy this equation:

p7 + q3 = r2, whenever gcd (p, q, r) = 1

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts re: sub-spoiler--another, via computer | Comment 2 of 3 |
(In reply to sub-spoiler by Ady TZIDON)

10   for R=1 to 1000000:T=R*R
20   for P=1 to int(T^(1/7))
30     for Q=1 to int((T-P^7)^(1/3)+0.5)
40       if P^7+Q^3=T then
50        :G=gcd(P,Q):G=gcd(G,R)
60        :print P,Q,R,G
70     next
80   next
90   next

finds

 p       q        r     gcd
 1       2       3       1
 3       9       54      3
 2       17      71      1
 8       64      1536    8
 8       128     2048    8
 9       243     4374    9
 10      225     4625    5
 9       486     10935   9
 15      225     13500   15

...

so (2,17,71) also fits


  Posted by Charlie on 2011-01-09 18:04:51
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 (7)
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