During the week 2-9 april, 2022 contributions came from Gennady Gusev,
Let p(n) - nth prime and p(n)# -
the product of all primes <= p(n).
By condition p(n)# = x*y and
Solving this system of equations
in integers for different primes we get a solution to the
p(n+1)=7, p(n)#=30: x=3,
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
I didn't find solution for other
primes < 50723.
I think there are no other
solutions at all.
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
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
7 = 2*5 - 3
11 = 3*7 - 2*5
13 = 5*11 - 2*3*7.