Problems & Puzzles: Conjectures

Conjecture 23. The Shyam's conjecture about the Rare Numbers

Since 1989, Shyam Sunder Gupta from India has been investigating the numbers RN (he names them "rare numbers") such that:

* RN + RN' = A2
* RN - RN' = B2
* RN' is RN reversed (or vice versa)
* RN RN' (RN is
not a palindrome)

Example: RN=65 because 65+56 = 11^2 & 65 - 56 = 3^2

The first 15 Rare Numbers RN are:

65, 621770, 281089082, 2022652202, 2042832002,868591084757, 872546974178, 872568754178, 6979302951885, 20313693904202, 20313839704202, 20331657922202, 20331875722202,20333875702202,  40313893704200, ... (See Neil's sequence A035519)

Shyam has demonstrated (*) that the RN's have the following properties:

Let RN =abcd...mnpq

Then a, b, p & q can be only one of the following alternatives, as a consequence of the digital properties of the squares:

Digital properties of RN

Property a q b & p
P1 2 2 b = p
P2 4 0 |b - p| = zero or even
P3 6 0 or 5 |b - p| = odd
P4 8 2 b + p = 9
P5 8 3 b - p=7 or p - b = 3
P6 8 7 b + p = 11 or b + p = 1
P7 8 8 b = p

From these properties the following corollaries can be stated:

Digital Corollaries of RN


 q can not be 1, 4, 6 & 9


  If q = 3 or 7, then the a = 8


 If a = q then (a = 2 or a = 8) and b = p


 a-q = 0 or 1 or 4 or 5 or 6


 Digital root of RN = 2 or 5 or 8 or 9

Currently Shyam has found a total of 63 RN numbers below 10^18, none of these ending in 3, five ending in 7 & none of these prime numbers.

Shyam wrote in his email dated the 4/2/2001:

It is now conjectured that there does not exist any Rare number which is also a Prime.
1) Can this conjecture be proved or disproved?
2) In case this is disproved, can anybody found a Rare Prime number?


I would add the following additional open questions about the RN numbers arising from the Shyam's study and results:

OQ1: Are the quantity of RN numbers infinite (**)?
OQ2: Find the first
RN ending in 3
OQ3: Can you develop an efficient algorithm
(***) to find the RN values? (****)

Other properties of RN (sent by Shyam the 15/2/2001) are the following:

1) A must be divisible by 11, if RN consist of even number of digits. So 121 must be a factor of A^2, if RN consist of even number of digits.

2) B must be divisible by 11 if RN consist of odd number of digits. Since B is always divisible by 3 , So B must be divisible by 33, if RN consist of odd number of digits. Hence 1089 must be a factor of B^2, if RN consist of odd number of digits.

It seems likely that odd RN's are fewer than even RN's. It is also likely that RN's with odd number of digits are fewer than RN's with even number of digits. So, odd RN's with odd number of digits are likely to still fewer. In fact only one RN i.e. 6979302951885 is known up to 10^18 which is odd and also consist of odd number of digits... In view of above , One additional Question: OQ4: Find the second odd RN which consist of odd number of digits?.


(*) Systematic computations of rare numbers, Shyam Sunder Gupta, The Mathematics Education, Vol. XXXII, No. 3, Sept. 1998.

(**) Palindrome Rare numbers are infinite. Shyam has proved this showing at least one particular example, that is to say a family of palindromes that are rare numbers: the first family is the palindromes 2(0)k4(0)k2 (242, 20402, 2004002, etc.)

(***) Shyam wrote in his email:  "I have developed a computer program in Fortran to calculate Rare numbers. In fact with refinement of the code over the years , the program has been made so powerful that all numbers up to 10^14 can be just checked for Rare numbers in less than a minute on Pentium III PC. In few hours I have been able to check up to 10^18."

(****) I would say that a radical improvement of the efficacy of new algorithms must come from the discovery of new/more digital properties of the RN's other than the published by Shyam. So an important task for the interested reader is to go deeper in the mathematical properties of these numbers.


Well, this conjecture is harder than I thought. As a matter of fact no one contribution has been done since posted. But Shyam continues working on the issue and got two more solutions, sent the 1/6/01:

"I am pleased to convey some new results which answers Q.4 of the Conjecture 23. The following two Rare Numbers (RN) of odd number of digits (i.e. 19 digits) are odd.


Both the above Rare Numbers are non-prime so the conjecture still holds good."


Shyam wrote on December 11, 2019:

OQ2: Find the first RN ending in 3

I am pleased to report that first Rare Number (RN)ending in 3 was found by me few years back but I missed
it to report earlier due to time constraints. The First RN ending in 3 consist of 22 digit.

RN = 8888070771864228883913 RN' = 3193888224681770708888
RN + RN' = 12081958996545999592801 = 109917964849^2
RN - RN' = 5694182547182458175025 = 75459807495^2

I also report that There are 124 Rare numbers below 10^22. All Rare numbers found below 10^22 are non-primes so the conjecture still holds good.


Metin Sariyar wrote on Dec 19, 2019:

I noticed that, All numbers m = a^2 + b^2 such that reversal(m) = 2*a*b is a term . For the numbers with this property, m-reversal(m) = (a-b)^2 and m + reversal(m) = (a+b)^2
Some terms with this property: 65, 621770, 2042832002, 872546974178, 872568754178, 20313693904202,... If it is proved that the numbers with this property are  infinitely many than the sequence is infinite. 
but for the sequence 
 R + R' = A3
 R - R'  = B
The sequence is infinite. Proof : The numbers of the form  R=3*10^2n+1 for n>=1  are terms (1030, 1000300, 1000003000, ... ) (but these may not be the all the terms of this seq.) then R'=10^3n+3*10^n for n>=1 . So, R-R'=10^3n+3*10^n-3*10^2n-1=(10^n-1)^3 (9^3,99^3,999^3,9999^3,99999^3,...)  and  R+R'=10^3n+3*10^n+3*10^2n+1=(10^n+1)^3 (11^3, 101^3, 1001^3,...)



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