Problems & Puzzles:
Recently I saw somewhere out there the
following nice puzzle: find numbers like these: N=A&B=A+B^3.
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:
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
is prime for some others k>3?
Q2. Are there prime solutions for
N=A&B=A+B^3 out of this family:
Contributions came from Ryan Bailey, J. K. Andersen & Emmanuel Vantieghem,
Jahangeer & Farideh.
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.
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) =
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.
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 >
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
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
Since each member of this family is an
even, there exist no prime numbers in this family of infinite
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 &
If B = 702, then A =
346294, A' = 347776 and we get N = 346294702
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 !