a prime is always next to a number that's divisible by six, like this:

n divisible by 2
n +1 divisible by 3
n +2 divisible by 2
n+ 3 prime number
n +4 divisible by 6
n +5 prime number
...
n + 6 divisible by 2
n + 7divisible by 3
[repeat]
and either the number divisible by 6 (n+4) or both the numbers before and after the set of prime numbers (n+2 and n+6)are divisible by 4.

also x² -1 = (x-1)(x+1)

So we effectively multiply the numbers under and above the prime number
As said they always have the factors "6" and "4" which multiplie give 24

if a prime number p is greater than 3 then p=3n+1 or 3n+2 ,n belongs to natural numbers, since its prime not divisible by 3 etc.

now p^2-1 =(3n+1)^2-1 = 9n^2 + 6n = 3n(3n+2)
or =(3n+2)^2-1 = 9n^2 + 12n + 3 = 3(3n^2+4n+1)

so in either case 3 divides p^2-1

also since its prime it can only be of one of the following forms in terms of divisibility by 4

4n+1, 4n+3 (4n +2 means divisible by 2)

now p^2 -1 =(4n+1)^2 -1 = 16n^2 +8n = 8n(2n+1)
or (4n+3)^2-1 = 16n^2 + 24n + 8 = 8(2n^2+3n+1)

so 8 divides p^2-1

now since 3 and 8 are co-prime gcd(3,8)=1 3*8= 24 divides p^2-1

I think there are more possible solutions CBA to find them all
the numbers are 4 and 13
Sam receives 17 and Pat 52

but all possible breakups of 17 =2+15=4+13=6+11=10+7=12+5=14+3

has ambiguity in product form in terms of grouping factors
2*15=2*3*5
4*13=2*2*13
8*11=2*2*2*11
10*7=2*5*7
12*5=2*2*3*5
14*3=2*7*3

so Sam knows Pat can not possibly know the numbers

now Pat sees 52=2*2*13 so its either 4,13 or 2,26
and so he says he doesnot know.
but if 2+26=28 was given to Sam atleast one of its breakups =11+17 is non ambiguous in product so Sam couldnot have made that earlier statement
So Pat is sure its not 2 and 26 and knows the number.

So now that Pat tells Sam he knows the numbers, Sam understands two things
1. Pat definitely had 2 possibilities and eliminated 1
so the product is one of 2*3*5,2*2*13,2*5*7,2*7*3
2. the eliminated product breakup has a sum breakup which leads to a non-ambiguous product.
(which would have prevented Sam's statement)

2*3*5=> 6+5=11 =2+9=4+7=6+5 all lead to ambiguous product
2*5*7=> 2+35=37 will have all ambiguous since its even+odd the factor 2 will create ambiguity
the corner case 2*35=2*5*7 is ambiguous too
2*7*3=> 2+21=23 same reasoning as above

only case is 2*2*13, where alternate break up 2+26=28 leads to a non-ambiguous breakup 11+17
so now Sam becomes sure this is indeed the case.
and numbers are 4 and 13.