Let denote the set of all integers (as usually it do ).
Consider a function with the following properties:
for all . Is it possible that all positive divisors of occur as values of ?
A happy note: is actually 1748 and it is written to retain symmetry in problem.
The answer is .
. . . (1)
. . . (2)
. . . (3)
Then, we have
. . . (4)
Again we have
. . . (5)
So, . . . .(6)
It’s your test of breaking and arranging numbers.
Similarly we can prove that by this way:
Therefore; .. …..(7)
This shows that is periodic by (the least we could get) and all possible values of are in the list . Finally by periodicity
In particular, .
Hence all possible values of function are and . In particular, assumes no more than 5 function values. However, 92 has 6 positive divisors 1, 2, 4, 23, 46 and 92. Hence, the all positive divisors of 92 cannot occur as values of . That’s why the answer is “No”.