> DivisorSigma[2k,m]/DivisorSigma[k,m] is integer

As Adam already disproved the "only if" part, let's concentrate on the

> if and only if

> m is a perfect square

>

> DivisorSigma[k,m]=sum[d^k, d divisor of m]

"if" one. First, there are two simple observations to make:

1) If p is a prime,

DivisorSigma[k, p^e] = 1 + p^k + ... + p^ek = (p^[k(e+1)] - 1)/(p^k-1)

2) If c and d are coprime,

DivisorSigma[k, c*d] = DivisorSigma[k, c] * DivisorSigma[k, d]

Thus, if m = p1^e2 * p2^e2 ... * pn^en is the prime factorization, we have

DivisorSigma[k, m] =

(p1^[k(e1+1)] - 1) / (p1^k - 1) *

(p2^[k(e2+1)] - 1) / (p2^k - 1) *

...

(pn^[k(en+1)] - 1) / (pn^k - 1).

and DivisorSigma[2k, m] =

(p1^[2k(e1+1)] - 1) / (p1^(2k) - 1) *

(p2^[2k(e2+1)] - 1) / (p2^(2k) - 1) *

...

(pn^[2k(en+1)] - 1) / (pn^(2k) - 1).

The ratio of these two is ten equal to

DivisorSigma[2k, m] / DivisorSigma[k, m] =

(p1^[k(e1+1)] + 1) / (p1^k + 1) *

(p2^[k(e2+1)] + 1) / (p2^k + 1) *

...

(pn^[k(en+1)] + 1) / (pn^k + 1).

Obviously, if all the exponents are even, each fraction of the product

is an integer (the numerator factors algebraically in such case). This

proves the "if" direction of your statement.

The smallest counterexample to the "only if" part seems to be m=20, k=1:

DivisorSigma[20, 2] = 1+4+16+25+100+400 = 546

DivisorSigma[20, 1] = 1+2+4+5+10+20 = 42

546 / 42 = 13

Peter

PS. Now I only have to hope that Yahoogroups do not mangle this mail

beyond comprehensibility :-)

--

[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278- Sebastián Martín Ruiz Said:

1. Divisor Function

Posted by: "Sebastián Martín Ruiz" sebi_sebi@... sebi_sebi

Date: Sun Aug 19, 2007 8:31 pm ((PDT))

Hello all:

I have obtained this result:

DivisorSigma[2k,m]/DivisorSigma[k,m] is integer

if and only if

m is a perfect square

DivisorSigma[k,m]=sum[d^k, d divisor of m]

Can anyone prove this result?

Sincerely

Sebastian Martin Ruiz

Kermit Said:

If m is a perfect square then

m = p1^a1 p2^a2 p3^a3 . . . pj^aj

where each of the a1, a2, .... aj are even.

If d divides m then

d = p1^b1 p2^b2 p3^b3 . . . pj^bj

where each b1,b2,...bj are less than or equal to the corresponding

a1,a2,...aj.

The

DivisorSigma[k,m]=sum[d^k, d divisor of m]

is product of

( 1 + p1^k + p1^(2k) + p1^(3k) + . . . + p1^(a1*k) )

(1 + p2^k + p2^(2k) + p2^(3k) + . . .+ p2^(a2*k) )

. . .

(1 + pj^k + pj^(2k) + pj^(3k) + . . . + pj^(aj * k) )

=

( p1^( a1 * k) - 1) / (p1^k - 1)

(p2^(a2 * k) - 1)/ ( p2^k - 1)

. . .

(pj^(aj*k) - 1) / (pj^k - 1)

When will this product be divisible by

( p1^( a1 * k/2) - 1) / (p1^k - 1)

(p2^(a2 * k/2) - 1)/ ( p2^k - 1)

. . .

(pj^(aj*k/2) - 1) / (pj^k - 1)

?

To see the answer more readily,

define a1 = 2 c1, a2 = 2 c2, ...aj = 2 cj

Then we are asking when is

[ p1^(2 c1 * k) -1 ) ( p2^(2 c2 * k) - 1 ) . . . ( 2 pj ^ ( 2 cj * k ) - 1) ]

/ [ p1^(c1*k) - 1) ( p2^(c2 * k) - 1 ) . . . ( 2 pj ^ ( cj * k ) - 1) ]

= [ p1 ^ (c1 * k) + 1 ) ( p2 ^ (c2 * k) + 1 ) . . . (pj * ( cj * k) + 1)

an integer?