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.

Note: these are spurious results based on use of insufficient precision (lack of DEFDBL A-Z) in the program.

4324  4233  4140            184           23
7196  5097  420             1232          17
5460  4393  2964            1872          3
7592  5409  936             832           41
4900  4803  4704            686           2

interpreted in the same fashion as before: the first three numbers in each row are the numerators; the next number is the common denominator; and at the end is the prime P.

Edited on March 27, 2013, 12:43 pm

Posted by Charlie on 2013-03-26 09:25:24