Problems & Puzzles: Puzzles

 Puzzle 908. Descending residues Vic Bold sent the following nice puzzle. For two consecutive primes, (p, q) such that q>p. q=nextprime(p) we construct the following scheme of descending residues: Let pq = p*q pq mod (p+q)=r1 pq mod r1=r2 pq mod r2=r3 ... pq mod rk-1= rk =1 (stop) or in a compact form: (p, q)-> {r1, r2, ...rk}, rk=1 Example by Vic: (p, q) = (11, 13), steps = 5 11*13 mod (11+13) = 23 11*13 mod 23 = 5 11*13 mod 5 = 3 11*13 mod 3 = 2 11*13 mod 2 = 1 (stop) or (11, 13) -> {23, 5, 3, 2, 1} In this example all the five residues are primes or 1. Carlos Rivera found a second example like the above such that steps = 6 and all the six residues are primes or 1: (53, 59) -> {103, 37, 19, 11, 3, 1} Rivera suggests that there are no larger scheme of pure primes or 1 residues, if p & q are consecutive primes. Q1. Can you find another pair of consecutive primes (p, q) with more that six residues prime or 1 and none composite?   Q2. Can you find another pair of primes (p, q) not necessarily consecutive, with more than six residues prime or 1 and none composite? Rivera has gotten a pair of consecutive primes (p, q) with 20 primes or 1, in a sequence of 40 descending residues: (3810860329, 3810860351) -> { 7621720559, 1905433679, 49252203, 30653789, 30177730, 20955199, 15150616, 10199407, 3078304, 2126007, 1455269, 1295988, 1132283, 674812, 553703, 449111, 278067, 156221, 79518, 42239, 12303, 6737, 2309, 1777, 1631, 719, 607, 281, 189, 131, 59, 48, 23, 17, 6, 5, 4, 3, 2, 1}; primes & 1 in the set are in bold type. Q3. Send another example of consecutive primes (p, q) with more than 20 residues prime or 1 and some composites.

Contributions came from Emmanuel Vantieghem and Fred Schneider

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Emmanuel wrote on Jan 19, 2018:

First of all, it should be observed that the sequence of descending residues not allways terminates in 1.
Sometimes it end in zero, for instance :

( 2, 7 )  ->  {5, 4, 2, 0}
( 3, 11 )  ->  {5, 3, 0}
( 7, 23 )  ->  {11, 7, 0}
I even think that for every prime  p  there exists a prime  q > p  such that  ( p, q )  ->  { ... , p , 0}.  But of course, this will be difficult to settle.
About the three questions you posed, this are my findings :
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Q1. I think Carlos is right, though there exists probably infinitely many primes  p  such that
( p, Nextprime(p) )  ->  a set of primes or 1.
The  p < 100000  are :  2, 3, 5, 7, 11, 13, 53, 179, 1019, 5639, 11699, 25799, 44699, 48479, 74099, 74759, 79559, 83339, 91139, 93239.

Q2.  My best result is : 11 primes when  p = 104827, q = 497411 :
( p, q )  ->  {336857, 7867, 4513, 2069, 773, 113, 103, 73, 11, 5, 2, 1}

Q3. Take  p = 10659800821 , q = nextprime(p).  Then :
( p, q )  ->  {21319601159, 5329969721, 3219495912, 2717977367, 564129422, 332515275, 258257732, 56242951, 35248152, 18462431, 13353620, 9755147, 9361289, 1674781, 580379, 273898, 255253, 226329, 140099, 107897, 102061, 86171, 67964, 43063, 26489, 10477, 3396, 971, 505, 487, 314, 239, 196, 59, 37, 17, 11, 10, 7, 3, 2, 1}, which contains the 21 primes :
21319601159, 56242951, 9755147, 580379, 255253, 107897, 102061, 86171, 43063, 26489, 10477, 971, 487, 239, 59, 37, 17, 11, 7, 3, 2

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Fred wrote on Jan 19, 2018:

Q2. This is the best solution (by far) that I have found:

Length = 14
788760464964743 = 3931787*200611189
26759 mod 204542976
353 mod 26759
199 mod 353
101 mod 199
37 mod 101
19 mod 37
17 mod 19
13 mod 17
11 mod 13
7 mod 11
5 mod 7
3 mod 5
2 mod 3
1 mod 2

I did this by setting up likely candidates with the Chinese Remainder Theorem, checking if a figure is a semi-prime and then checking the residues.

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