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 p^{2} = a.(a+1)
Q1. Find the next 3
primes of this type [p^{2}
= a.(a+1)]
Q2. Find a titanic
prime of this type [p^{2}
= a.(a+1)].
Solution:
Solutions to Q1 & Q2 were sent by
J. C. Rosa and Patrick de Geest, respectively.
***
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*90911
p=k*2161*9091+1
p=k*2161*9091+701*90911
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 digitrelated
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 " Neartautonymic 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 JeanClaude 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
'3PRP' are respectively 8217 and 9291 and so are almost gigantic ones ! I
shall try to find a genuine gigantic probable prime solution.
***
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.
***
