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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    
Rating: 3.0000 (1 votes)

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: still no luck on part 2broll2013-03-29 10:09:08
re(2): more results, but still no part 2 solnCharlie2013-03-27 11:29:40
re: more results, but still no part 2 solnbroll2013-03-27 09:08:51
Hints/Tipsre: still no luck on part 2broll2013-03-27 08:02:19
re: still no luck on part 2broll2013-03-27 04:49:33
Some Thoughtsstill no luck on part 2Charlie2013-03-26 14:21:36
Some Thoughtsmore results, but still no part 2 solnCharlie2013-03-26 09:25:24
Some Thoughtspart 1b. (spoiler)Charlie2013-03-26 08:46:42
Some Thoughtspart 1a (spoiler)Charlie2013-03-26 08:35:08
re: Are there solutions?broll2013-03-26 01:11:09
Are there solutions?Jer2013-03-25 22:08:49
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