- --- jbrennen <jb@...> wrote:
> What is the smallest N such that sigma(N) is congruent to

Well, an upper bound is 160470643909878751793805444097921

> 17 (modulo 102)?

Phil

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--- In primenumbers@yahoogroups.com, "jbrennen" <jb@...> wrote:

> What is the smallest N such that sigma(N) is congruent to

> 17 (modulo 102)?

An upper bound is 160470643909878751793805444097921

Phil - I think you can prove that Phil's example 103^16 is the smallest

such number. Sigma(n) has factors (1+p+p^2+...+p^k) for p^k||n (<-

largest power of p that divides n). For sigma(n) to be 5 mod 6, all

the factors have to be either 1 or 5 mod 6 (i.e., no factors that

are 2,3,4,0 mod 6). By examination, for p=2 and p=3,

(1+p+p^2+...+p^k) is never 0 mod 17 and 1 or 5 mod 6.

For p=3, we get 1+3+3^2+...+3^k odd when k is even but then

1+3+3^2+...+3^k=(3^(k+1)-1)/2 while the order of 3 mod 17 is 16 so 3^

(k+1)-1 is divisible by 17 only when 16 divides k+1. It can not be

true that k is even and 16 divides k+1.

For p=2, we get 1+2+2^2+...+2^k is 1 mod 3 when k is even. But the

order of 2 mod 17 is 8, so 2^(k+1)-1 == 0 mod 17 means (k+1) is even.

The only other available primes are either 1 or 5 mod 6. If p is 5

mod 6 and 1+p+...+p^k is either 1 or 5 mod 6, then k is even. If p

is 1 mod 6 and 1+p+...+p^k is either 1 or 5 mod 6, then k is either

4 or 6 mod 6.

For k even, solve 1+x+...+x^k==0 mod 17 and find the first solution

happens when k=16 and x==1 mod 17. The first prime that is 1 mod 17

is 103. 103^16 is the first number that solve sigma(n)==17 mod 102. - --- In primenumbers@yahoogroups.com, "Adam" <a_math_guy@...> wrote:
>

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

> I think you can prove that Phil's example 103^16 is the smallest

> such number.

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.