Problems & Puzzles: Puzzles

Puzzle 257.  Primes and sibling numbers

The prime 727 is the first prime such that its square 528529 is a sibling number of the type a.(a+1) where a=528 and here the dot "." means concatenation.

So, in this puzzle we are interested in primes p such that p2 = a.(a+1)

Q1. Find the next 3 primes of this type [p2 = a.(a+1)]

Q2. Find a titanic prime of this type [p2 = a.(a+1)].

Solution:

Solutions to Q1 & Q2 were sent by J. C. Rosa and Patrick de Geest, respectively.

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Q1. J. C. Rosa has found the first eight primes of the asked type:

727, 37380894211, 9090909090911, 827322055537823, 940019607538439, 71037310558865507, 586677551819414769713, 653061224489795918369 and no other prime solution up to 10^25

BTW, his solution 37380894211was already published in the Plate 152 of the well known Patrick de Geest's site, World of Numbers.

This is the JCR's method:

We want p such that p^2=a.(a+1) that is to say : p^2=a*10^L+a+1 from where : p^2=a*(10^L+1)+1 (1) where L=length of (a+1)=length of a=length of p.

Let d a prime divisor of (10^L+1). The equality (1) means that : p^2=1 mod d and therefore : p=+/-1 mod d . Thus p=k*d+/-1 ( p and d are odd number therefore k is even )

For a quick search , take for d the greatest prime divisor of (10^L+1).

Examples:

1°) L=13 . 10^13+1=11*859*1058313049

d=1058313049 , p=1058313049*k+/-1 length of p=13 implies : 10^12<p<10^13 , from where:944<k<9450 start from k=944 to k=9450 step 2 we obtain only 2 solutions:

p=6100616414437 and p=9090909090911 ( prime)

2°) L=15 , 10^15+1=7*11*13*211*241*2161*9091.

Here the prime factors are small. Take the two greatest : 2161 and 9091. Starting from the equalities : p=+/-1 mod 2161 and p=+/-1 mod 9091, after some calculations we obtain , for p, four forms:

p=k*2161*9091-1
p=k*2161*9091+1
p=k*2161*9091+701*9091-1
p=k*2161*9091+1460*9091+1.

We obtain many solutions but only two primes:

p=827322055537823 , p=9400019607538439.

Q2. Patrick de Geest sent the following solutions:

What you named in your puzzle as 'sibling numbers' is known to me as tautonymic numbers. I studied these digit-related numbers a while ago and collected what I had in a wonplate (see http://www.worldofnumbers.com/em152).

Also Jean Claude Rosa made an important contribution to the topic.

In the subsection " Near-tautonymic numbers of the form (T)_(T + 1) as SQUARES " one can find the first two primes as asked in question Q1 :

[ 727 ]^2 = 528_529
[ 37380894211 ]^2 = 13973312520_13973312521

Unfortunately the list ends before the third and fourth prime showed up. No doubt Jean-Claude Rosa will give the next few terms.

What does show up in the list is a pattern starting from the given example 727 :

[ (6n)7(3n)2(6n)7 ]^2 = (4n)5(3n)2(8n)8_(4n)5(3n)2(8n)9

(6n) means '6' repeated n times

By putting the rootnumber in an ABC2 file for PFGW I was able to find some n values yielding probable primes. I hope this suffices as a solution to question Q2 of your puzzle. The puzzle doesn't ask for the smallest titanic solution only "Find 'a' titanic prime...". primes for the following n values = 0, 16, 17, 33. probable primes for n = 2738, 3096

Note that the digit lengths of these last two '3-PRP' are respectively 8217 and 9291 and so are almost gigantic ones ! I shall try to find a genuine gigantic probable prime solution.

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Phil Carmody found (March 2003) the earliest (supposedly) titanic prime of the asked type (only the first 10 initial and 10 ending digits shown, CR; the whole prime can be sent on request)

5107962136...9079192809, 1000 digits

See his method described in the Puzzle 258.

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