Problems & Puzzles: Puzzles

 Puzzle 1081 2*7*13-3*5*11=17 JM Bergot sent the following nice puzzle Take all the primes less than 17 and using each one just once create two numbers such that their difference equals 17: 2*7*13=182 and 3*5*11=165 giving a difference of 17. Q. Can this be done with a larger prime and all the primes less than this prime?

During the week 2-9 april, 2022 contributions came from Gennady Gusev, Oscar Volpatti

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Let p(n) - nth prime and p(n)# - the product of all primes <= p(n).
By condition p(n)# = x*y and y-x=p(n+1).
Solving this system of equations in integers for different primes we get a solution to the problem.
p(n+1)=7, p(n)#=30: x=3, y=10=2*5 ->2*5-3=7
p(n+1)=11, p(n)#=210: x=10, y=21 -> 3*7-2*5=11
p(n+1)=13, p(n)#=2310: x=42, y=55 -> 5*11-2*3*7=13
p(n+1)=17, p(n)#=30030: x=165, y=182 (see condition of the puzzle).

I didn't find solution for other primes < 50723.
I think there are no other solutions at all.

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Oscar wrote:

There are no solutions for primes p between 17 and 10^7.

Given a target prime p > 2, let q be the product of all primes below p; we need to split q into two factors x and y, such that the difference x-y equals p itself.
Consider the numbers x and -y; their sum is p and their product is -q, so they are the roots of the following quadratic equation:
z^2 - p*z - q.
Its discriminant d = p^2+4*q is always an odd positive integer, so the roots are always real and distinct; the solution formula involves parameter r = √d:
x = (r+p)/2
y = (r-p)/2
If d is a perfect square, then r is an odd integer, x and y are integers with opposite parity; otherwise r is irrational, x and y are irrational too.

I computed d for all primes p < 10^7, finding perfect squares only for 5 <= p <= 17.
Here are the small resulting solutions, easy to find by hand:
5 = 2*3 - 1 (maybe illegal, y empty product?)
7 = 2*5 - 3
11 = 3*7 - 2*5
13 = 5*11 - 2*3*7.

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