## Re: Prime chains x-->Ax+B [puzzle 11]

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• ... Puzzle 11: Find a plupluperfect chain of 12, i.e. a triplet (A, B, p[1]) such that A = 1 mod 11, B is coprime to 11, and the iteration p[n+1] = A*p[n] + B
Message 1 of 143 , Dec 1, 2010

> Here's a plupluperfect chain of 8:
>
> p=3;for(k=1,9,print([k,factor(p)]);p=140820*p-422453)
> [1, Mat([3, 1])]
> [2, Mat([7, 1])]
> [3, Mat([563287, 1])]
> [4, Mat([79321652887, 1])]
> [5, Mat([11170075159124887, 1])]
> [6, Mat([1572969983907966164887, 1])]
> [7, Mat([221505633133919795338964887, 1])]
> [8, Mat([31192423257918585579633034964887, 1])]
> [9, [7, 1; 627502434740013603046274854822137841, 1]] \\ Bounded!

Puzzle 11: Find a plupluperfect chain of 12, i.e. a triplet
(A, B, p[1]) such that A = 1 mod 11, B is coprime to 11,
and the iteration p[n+1] = A*p[n] + B yields a chain of
12 increasing primes, p[n], for n = 1 to 12.

Comments: Clearly, p[2] = 11 and hence B = 11 - A*p[1], with
p[1] = 2, 3, 5 or 7. So this boils down to 4 sub-problems of
Carmody-scale difficulty (maybe easy for Phil; harder for mere
mortals like Kevin, Mike, or me). I did not yet find a solution.

David
• ... Suppose that we want to start with a square and get a square. Then we must solve the Diophantine equation y^2 = a*x^2 + b For any pair (a,b), Dario will
Message 143 of 143 , Jan 7, 2011