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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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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):

y=41
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
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