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

Home > Numbers
Conga Primes (Posted on 2013-03-23) Difficulty: 3 of 5

x^2-y^2 = y^2-z^2 = 5 is a classic problem that can be solved in the rationals, with, e.g.:

(49/12)2-(41/12)2 = (41/12)2-(31/12)2 = 5 (Fibonacci).

We seek non-trivial rational solutions to x^2-y^2 = y^2-z^2 = P, with P prime. Since we can always find compound multiples of such solutions with other primes happily joining the chain, let's call these paragons 'conga primes'. (Conversely, primes that only appear in conjunction with other primes could be 'tango primes', since it takes at least two...)

1. Solve over the rationals:
x^2-y^2 = y^2-z^2 = 7
x^2-y^2 = y^2-z^2 = 41

2. Give an example of a 'conga prime', P, greater than 41, such that x^2-y^2 = y^2-z^2 = P.

See The Solution Submitted by broll    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Are there solutions? | Comment 1 of 11
I can't find one brute force with numerator of the middle number up to 5000.  Have I not searched far enough?

If you want to laugh at my programming (Haven't used basic in 20+ years):

for a = 5 to 5000
for b = 1 to (a-1)
    C = sqr((a^2-b^2)/y)
    D = sqr(5*C^2+a^2)
    IF C=int(C) and D = int(D) then print a, b, C, D
next b
print a
next a

  Posted by Jer on 2013-03-25 22:08:49
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information