Problems & Puzzles: Puzzles

 Puzzle 642. N=A&B=A+B^3 Recently I saw somewhere out there the following nice puzzle: find numbers like these: N=A&B=A+B^3. Two examples: 568 = 56 + 8^3 86044 = 860 + 44^3 Are there primes of this type? Well, I found that there is an family of infinite solutions of this form: (9)k-18(0)k(9)k , for k=1, 2, 3, .... such that is prime for k=1 & k=3, that is to say for 809 & 998000999. Q1. Can you check if (9)k-18(0)k(9)k is prime for some others k>3? Q2. Are there prime solutions for N=A&B=A+B^3 out of this family: (9)k-18(0)k(9)k?

Contributions came from Ryan Bailey, J. K. Andersen & Emmanuel Vantieghem, Jahangeer & Farideh.

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Ryan wrote:

Q1. I found that for k = 637 & 839 the number corresponding to k in the sequence is prime.
Q2. I checked each combination of String permutations for all primes from 1 to 5 X 10^9, and found no new primes that satisfied such condition, not to say there aren't any out of that range.

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Jens wrote:

Q1. PrimeForm/GW has found and proved primes for k = 637, 839, 2197. There are no other primes for k < 25000.

Q2. There are no prime solutions out of the family. All other solutions have a
common factor in A and B, so this will be a factor in A+B^3. Proof:
If B has k digits then we have A*10^k+B = A+B^3.
This can be written A*(10^k-1) = B*(B-1)*(B+1).
The only way to get coprime A and B is if B = 10^k-1.
Then A = (B-1)*(B+1) = (10^k-2)*10^k, and we get the known family of solutions.

All other (composite) solutions for a given k can be generated by
factoring 10^k-1 and using the Chinese remainder theorem to compute
the value of B for each way to split the factors between B, B-1 and B+1.

***

Emmanuel wrote:

Q1.  There are two more values for  k  that give a prime :

k = 637, giving  9...(636 nines)...980...(637 zeros)...09...(637 nines)...9 ; 1911 digits certified prime by PRIMO in 4h 04' 21"

k = 839, giving   9...(838 nines)...980...(839 zeros)...09...(839 nines)...9 ; 2517 digits certified prime by PRIMO in 17h 44' 38"

If there is still another  k  that gives a prime N, then  k > 1000.

Q2. If  A, B  satisfy  A&B = A+B^3 = N, then : B^3 - B = A(10^k - 1),    (*)

where  k  is the number of digits of  B.

So, to find all possible  N  we simply search for all  k-digit  B  for which the expression  B^3 - B  is divisible by  10^k - 1.  We then get  A as the quotient of  B^3 - B  and  10^k - 1.

Now, let  A  and  B  satisfie  (*).  If  p  is a prime divisor of  B  but not of  10^k - 1,  then  p  will divide  A  and the number  N = A&B  is divisible by  p.  Thus, when  N  is prime, B will be a divisor of  10^k - 1.  Since  B  must have  k  digits, and since  (*)  must hold, there are only two possibilities to get a prime  N :

1)  B = 9...9  (k  nines)  which leads to the solutions in Q1
2)  B = 9...9/7 (6n  nines) = (10^(6n) - 1)/7 = (concatenation of  n times  142857).

In the second case, we have

N = B + (B^3 - B)10^(6n)/(10^(6n) - 1)

= ((B^2)(10^(6n)) - 1)/7

= (B*10^(3n) - 1)*(B*10^(3n) + 1)/7.

This is an integer iff  n = 7w+1  or  7w - 1.  In neither case  N  is a prime !That means that only the 9-8-0-9-family can deliver prime

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Jahangeer & Farideh wrote:

Answer to Q1 : (9)k-18(0)k(9)k  is probable prime for k = 637,839 & 2197.

...

Consider numbers of the form  (9)k-15(0)k-1 (9)k-168 is another family of infinite solutions for N. The first example 568 = 56+8^3 is obtained for k=1.
Since each member of this family is an even, there exist no prime numbers in this family of infinite solutions.

It is interesting that for some numbers B,  B = 8, 44, 54, 98, 296, 702, 998, ...
both numbers A+B^3 & A'+(B+1)^3 are solutions for N.

If B = 8, then  A = 56, A' = 80 and we get N = 568 & N = 809.
If B = 44, then  A = 860, A' = 920 and we get N = 86044 & N = 92045.
If B = 54, then  A = 1590, A' = 1680 and we get N = 159054 & N = 168055.
If B = 98, then  A = 9506, A' = 9800 and we get N = 950698 & N = 980099.
If B = 296, then  A = 25960, A' = 26224 and we get N = 25960296 & N=26224297.
If B = 702, then  A = 346294, A' = 347776 and we get N = 346294702 & N=347776703.
If B = 998, then  A = 995006, A' = 998000 and we get N = 995006998 & N = 998000999

...

We haven't found the relation between A & A', yet !

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