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.

Charlie,

You might be able to speed things up by looking for xy^3-x^3y = (6a)^2(P), with P the desired prime. Then 4*(xy^3-x^3y) = d will provide a difference: b^2-a^2 = c^2-b^2 = d, generating a,b,c.

This might be easier than scanning thousands of differences of squares for a match!