## Re: Puzzle - Law of small numbers

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• ... Yes, and you seemed to give a convincing argument. :) Did anyone notice that I basically gave away the key to finding Phil s example? I stated that
Message 1 of 5 , Jan 10 3:25 PM
>
> I think you can prove that Phil's example 103^16 is the smallest
> such number.

Yes, and you seemed to give a convincing argument. :)

Did anyone notice that I basically gave away the key to finding
Phil's example? I stated that solving sigma(N) == 5 (mod 6)
was easily done by picking a prime p of the form 6*x+1 and
then taking p^4, or p^(5-1).

This is of course easily generalized...

So to solve sigma(N) == 17 (mod 102), find a prime of the
form 102*x+1. The first such is of course 103. And then
set N = 103^(17-1) = 103^16.

The tricky part is proving that 103^16 is in fact the minimal
solution.

My method for doing so:

Note that N must be divisible by a prime power p^a which
has sigma(p^a) divisible by 17 but not divisible by 2 or 3.

Note also that sigma(p^a) = (p^(a+1)-1)/(p-1).

In order for sigma(p^a) to be an odd number (not divisible
by 2), we need to have either p or a be even.

If p is even, it must be 2, and we would have:
p^(a+1) == 1 (mod 17). It turns out that this is true
when a == 7 (mod 8); however, when a == 7 (mod 8), we
also have sigma(2^a) divisible by 3. So that doesn't
work.

So p must be odd and a must be even. In order for
p^(a+1)-1 (the numerator of the expression above) to
be divisible by 17, (a+1) must be divisible by the
order of p modulo 17. Note that the order of p
modulo 17 must divide evenly into 16 (a consequence
of Fermat's Little Theorem). So we have an
odd number (a+1) which must be divisible by a
divisor of 16. Clearly the divisor (the order of
p modulo 17) must be 1. This implies that the
prime p is in fact of the form 17*x+1.

And once we know that p == 1 (mod 17), we can see
easily that a must be of the form 17*y-1.

The smallest prime p of the form 17*x+1 is 103,
and the smallest exponent a of the form 17*y-1 is 16.

So the smallest prime power which has sigma(p^a)
divisible by 17, but not by 2 or 3, is 103^16.

So any solution N to the original puzzle must be
divisible by a prime power >= 103^16.

There are not a lot of ordered pairs (A,B) such
that A > 2 and such that N = (B+1)^(A-1) is the
smallest solution to sigma(N) == A (modulo B).

I know of these three: (5,6), (5,30), and (17,102).

There could be others, if anyone wants to try
to find them. It's interesting that for those
three I know of, A is always a Fermat prime.
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