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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    
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re(2): more results, but still no part 2 soln | Comment 10 of 11 |
(In reply to re: more results, but still no part 2 soln by broll)

"Strictly speaking, e.g. (4324/184)^2 - (4233/184)^2 = 778687/33856, while (4233/184)^2-(4140/184)^2 = 778689/33856, with neither integral and a tiny difference of 1/16928. "

That's what I get for leaving out the DEFDBL A-Z. I wish QB defaulted to that.

  Posted by Charlie on 2013-03-27 11:29:40
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