Problems & Puzzles: Puzzles

Puzzle 116. A=B+C, A*B*C is a primorial...

Naohiro Nomoto asks for numbers A, B & C such that the following four rules apply:

  1. A = B + C
  2. A*B*C is a primorial
  3. A, B, & C are, each, a product of n of distinct primes
  4. gcd(A,B)=gcd(A,C)=1

Nomoto has found only two examples:

5 = 2 + 3

1105 = 231 + 874 [ 5*13*17 = 3*7*11 + 2*19*23 ]

Question 1: Find the next example

If the condition 3 is relaxed and we let that A, B & C are the product of distinct primes, not necessarily the same quantity, example:

10 = 3 + 7  [ 2*5 = 3 + 7 ]

There are some more examples, being the largest currently found, the following one:

496961=495726+1235 [ 17x23x31x41 = 2x3x7x11x29x37 + 5x13x19 ]

Question 2. Find the next 3 examples

Question 3. Can you argue if this sequence is finite or infinite?

Notes:  All the numbers for the sequence described in the question 2 are a subset of the
Sloane's integers sequence A057035.

Hakan Summakoğlu wrote (May 2011)

Q2: 219965 = 213486 + 6479  [ 5x29x37x41 = 2x3x7x13x17x23 + 11x19x31]


Jan van Delden wrote (May 2011)

I was wondering about the number of solutions of Puzzle 116, I thought that could be improved. Must have missed this one..

My contribution:


We have P:=p[n]#=a.b.c and a=b+c. Given P there are two equations with 3 unknowns. We can therefore (for instance) choose b and a and c follow. From this one can deduce that if b^2+4P/b is a square, say D^2, we have: a=(D+b)/2 and c=(D-b)/2.

Furthermore the minimum number of factors of P to be assigned to a can be calculated beforehand (using the largest prime factors of P), leading to the maximum number of factors to be assigned to b. Unfortunately this maximum is about n/3, which still gives a lot of tests to perform!

I found the following 20 solutions, in order of appearance (primarily Q2):

5=2+3 (Q1)

10=3+7 2x5=3+7

35=  2+33 5x7=2+3x11
22=  7+15 2x11=7+3x5
21=11+10 3x7=11+2x5

55=13+42 5x11=13+2x3x7

221=11+210 13x17=11+2x3x5x7
182=17+165 2x7x13=17+3x5x11
231=10+221 3x7x11=2x5+13x17

442=57+385  2x13x17=3x19+5x7x11
357=247+110 3x7x17=13x19+2x5x11

1430=119+1311 2x5x11x13=7x17+3x19x23
1105=874+231  5x13x17=2x19x23+3x7x11 (Q1)

2958=1495+1463  2x3x17x29=5x13x23+7x11x19

23345=374+22971 5x7x23x29=2x11x17+3x13x19x31

496961=1235+495726  17x23x31x41=5x13x19+2x3x7x11x29x37
219965=6479+213486  5x29x37x41=11x19x31+2x3x7x13x17x23
144210=16523+127687 2x3x5x11x19x23=13x31x41+7x17x29x37
226083=6118+219965  3x11x13x17x31=2x7x19x23+5x29x37x41
106981=57646+49335  7x17x29x31=2x19x37x41+3x5x11x13x23

No further solutions with n<=35


I computed the minimum absolute distance between b^2+4P/b and the nearest square (where no solution was found), let's call this dist(D^2). The smallest distance 103 was found (n in [12,14-26]) at n=18, b=12509182. At first sight this is large, however a and c fail to be integer by a mere 0.000000133 (=dist(a),dist(b)). However the resulting distance between P and a.b.c is huge, since a.b.c=(D+b)/2.b.(D-b)/2=b.(D^2-b^2)/4, this gap is of the order b.dist(D^2)/4=12509182.103/4.

In general dist(a)=dist(b)=dist(D^2)/(4D) and dist(P)=b.dist(D^2)/4. And dist(D^2) is maximal D+1/4. More should be known about the distribution of dist(D^2) to say more.





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