Distinct perfect squares with equal abundancy indices

[joseabdris]

When do distinct perfect squares give rise to equal abundancy indices? Specifically, when would it be true that:

sigma(A^2)/A^2 == sigma(B^2)/B^2

if A^2 is NOT equal to B^2 ???

It might help to know that I am only considering those A and B for which:

8/5 < sigma(A^2)/A^2 < 2 and 8/5 < sigma(B^2)/B^2 < 2.

Any helpful comments on this problem would be greatly appreciated.

[Gerry Myerson]

In article
<22151620.1148906131...@nitrogen.mathforum.org>,
joseabdris <josea...@yahoo.com> wrote:

> When do distinct perfect squares give rise to equal abundancy indices?
> Specifically, when would it be true that:
>
> sigma(A^2)/A^2 == sigma(B^2)/B^2
>
> if A^2 is NOT equal to B^2 ???

I don't know. Do you have any examples?

--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)

[joseabdris]

> In article
> <22151620.1148906131...@nitrogen.math

> forum.org>,
> joseabdris <josea...@yahoo.com> wrote:
>
> > When do distinct perfect squares give rise to equal
> > abundancy indices?
> > Specifically, when would it be true that:
> >
> > sigma(A^2)/A^2 == sigma(B^2)/B^2
> >
> > if A^2 is NOT equal to B^2 ???
>
> I don't know. Do you have any examples?
>
> --
> Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for
> email)

I don't have any specific examples. What I do know is that if sigma(A^2)/A^2 == sigma(B^2)/B^2 and A^2 is NOT equal to B^2 (with the added conditions that 8/5 < sigma(A^2)/A^2 < 2 and 8/5 < sigma(B^2)/B^2 < 2), then the following are true:

(1) sigma(A^2) is NOT equal to sigma(B^2).
(2) B^2 divides (A^2 sigma(B^2)). (Of course, B^2 does not divide sigma(B^2) but it does not follow from these that B^2 divides A^2.)
(3) A^2 divides (B^2 sigma(A^2)). (Again, A^2 does not divide sigma(A^2) but it does not follow from these that A^2 divides B^2.)

[Robert Israel]

In article <gerry-C7D7B2....@sunb.ocs.mq.edu.au>,

Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> wrote:
>In article
><22151620.1148906131...@nitrogen.mathforum.org>,
> joseabdris <josea...@yahoo.com> wrote:
>
>> When do distinct perfect squares give rise to equal abundancy indices?
>> Specifically, when would it be true that:
>>
>> sigma(A^2)/A^2 == sigma(B^2)/B^2
>>
>> if A^2 is NOT equal to B^2 ???
>
>I don't know. Do you have any examples?

There are none for 1 <= A < B <= 300000, according to Maple.

Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

[joseabdris]

Indeed, without loss of generality, assume that A < B with sigma(A^2)/A^2 = sigma(B^2)/B^2.

Let A = PROD\(i=1 to i=R) [(p_i)^(a_i)] and B = PROD\(i=1 to i=S) [(q_i)^(b_i)] be the prime factorizations of A and B, respectively.

Then A^2 = PROD\(i=1 to i=R) [(p_i)^2(a_i)] and B^2 = PROD\(i=1 to i=S) [(q_i)^2(b_i)].

Consequently:

sigma(A^2) = PROD\(i=1 to i=R) [sigma((p_i)^2(a_i))]
sigma(B^2) = PROD\(i=1 to i=S) [sigma((q_i)^2(b_i))]

since the sigma function is multiplicative.

Hence:

sigma(A^2) = PROD\(i=1 to i=R) [SUM\(j=0 to j=2(a_i) [(p_i)^j]]
sigma(B^2) = PROD\(i=1 to i=S) [SUM\(j=0 to j=2(b_i) [(q_i)^j]]

by the definition of the sigma function.

Suppose R = S, p_i == q_i and a_i == b_i for all i = 1, 2, ..., R=S.

Then A = PROD\(i=1 to i=R) [(p_i)^(a_i)] = PROD\(i=1 to i=S) [(q_i)^(b_i)] = B. A contradiction.

Therefore:

(1) R is not equal to S; OR
(2) There exists at least one i such that p_i is NOT equal to q_i; OR
(3) There exists at least one i such that a_i is NOT equal to b_i.

Hence, to prove that there are NO distinct perfect squares with equal abundancy indices, one needs to show that a contradiction results from assuming each of the 3 cases above and the additional condition that:

PROD\(i=1 to i=R) [SUM\(j=0 to j=2(a_i) [(p_i)^j]]/PROD\(i=1 to i=R) [(p_i)^2(a_i)] == PROD\(i=1 to i=S) [SUM\(j=0 to j=2(b_i) [(q_i)^j]]/PROD\(i=1 to i=S) [(q_i)^2(b_i)]

This is as far as I can go. It would seem that case 1 is the hardest one to tackle at this point.

[joseabdris]

If one could prove that:

gcd(A^2, sigma(A^2)) = gcd(B^2, sigma(B^2)) = 1, (***)

then it would follow that A^2 divides B^2 and also that B^2 divides A^2. From these two, it would follow that A^2 = B^2, a contradiction. Hence, it should follow that:

CONJECTURE: Distinct perfect squares give rise to distinct abundancy indices.

Can one use the fact that:

8/5 < sigma(A^2)/A^2 < 2 and 8/5 < sigma(B^2)/B^2 < 2

to prove (***)?

Would it be of any help to prove (***) or the Conjecture above if we restrict A and B to be odd?

[Gerry Myerson]

In article
<12825004.1149516771...@nitrogen.mathforum.org>,
joseabdris <josea...@yahoo.com> wrote:

> If one could prove that:
>
> gcd(A^2, sigma(A^2)) = gcd(B^2, sigma(B^2)) = 1, (***)

It is, in general, not true that gcd(A^2, sigma(A^2)) = 1,
e.g., check out A = 39.

> Would it be of any help to prove (***) or the Conjecture above if we restrict
> A and B to be odd?

In the example I've given, A is odd.

[joseabdris]

> In article
> <12825004.1149516771...@nitrogen.math

> forum.org>,
> joseabdris <josea...@yahoo.com> wrote:
>
> > If one could prove that:
> >
> > gcd(A^2, sigma(A^2)) = gcd(B^2, sigma(B^2)) = 1,
> (***)
>
> It is, in general, not true that gcd(A^2, sigma(A^2))
> = 1,
> e.g., check out A = 39.
>
> > Would it be of any help to prove (***) or the
> Conjecture above if we restrict
> > A and B to be odd?
>
> In the example I've given, A is odd.
>
> --
> Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for
> email)

Thank you for providing that counterexample, Gerry.

However, as I had previously posted, I require that:

8/5 < sigma(A^2)/A^2 < 2. In your example, A = 39 = 3*13, hence sigma(A^2) = 3*13*61. Therefore:

sigma(A^2)/A^2 = sigma((39)^2)/(39)^2 = 61/39 < 8/5.

Can you give me an example satisfying:

(1) gcd(A^2, sigma(A^2)) > 1
(2) 8/5 < sigma(A^2)/A^2 < 2 ??

[joseabdris]

Suppose gcd(A^2, B^2) = 1, then from (2) and (3) above, this implies that A^2 divides sigma(A^2) and also that B^2 divides sigma(B^2), which are both contradictions. Hence, gcd(A^2, B^2) >= 2. If we restrict A and B to be odd, gcd(A^2, B^2) >= 3.

Suppose gcd(A^2, B^2) = k, k >= 3. Then k divides A^2, and k also divides B^2, which gives rise to the equations:

(a) A^2 = kx, for some positive integer x.
(b) B^2 = ky, for some positive integer y.

Does it follow that k is a perfect square?

[Robert Israel]

In article <1542984.11496530948...@nitrogen.mathforum.org>,

joseabdris <josea...@yahoo.com> wrote:
>
>Can you give me an example satisfying:
>
>(1) gcd(A^2, sigma(A^2)) > 1
>(2) 8/5 < sigma(A^2)/A^2 < 2 ??

A = 21.

[Gerry Myerson]

In article
<20061203.1149653947...@nitrogen.mathforum.org>,
joseabdris <josea...@yahoo.com> wrote:

> Suppose gcd(A^2, B^2) = k, k >= 3. Then k divides A^2, and k also divides
> B^2, which gives rise to the equations:
>
> (a) A^2 = kx, for some positive integer x.
> (b) B^2 = ky, for some positive integer y.
>
> Does it follow that k is a perfect square?

gcd(A^2, B^2) = (gcd(A, B))^2.

[joseabdris]

> In article
> <20061203.1149653947...@nitrogen.math

> forum.org>,
> joseabdris <josea...@yahoo.com> wrote:
>
> > Suppose gcd(A^2, B^2) = k, k >= 3. Then k divides
> A^2, and k also divides
> > B^2, which gives rise to the equations:
> >
> > (a) A^2 = kx, for some positive integer x.
> > (b) B^2 = ky, for some positive integer y.
> >
> > Does it follow that k is a perfect square?
>
> gcd(A^2, B^2) = (gcd(A, B))^2.
>
> --
> Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for
> email)

So, if we let gcd(A, B) = k, then k >= 3 (since A and B are odd) and gcd(A^2, B^2) >= 9.

Since gcd(A^2, sigma(A^2)) = gcd(B^2, sigma(B^2)) = 1 implies A^2 divides B^2 and B^2 divides A^2 (since sigma(A^2)/A^2 = sigma(B^2)/B^2) and therefore A^2 = B^2 (a contradiction), then gcd(A^2, sigma(A^2)) > 1 OR gcd(B^2, sigma(B^2)) > 1.

How can one use these facts, then, together with 8/5 < sigma(A^2)/A^2 = sigma(B^2)/B^2 < 2, to produce some numerical information about A^2 and B^2?

[joseabdris]

Suppose that, using additional conditions, I could prove the stronger statement:

gcd(A^2, sigma(A^2)) > 1 AND gcd(B^2, sigma(B^2)) > 1.

Let d_0 = gcd(A^2, B^2). (Note that d_0 is an odd perfect square; therefore d_0 >= 9.) Then d_0 is the largest among the common divisors of A^2 and B^2. Thus, if we let d_1 = gcd(A^2, sigma(A^2)) and d_2 = gcd(B^2, sigma(B^2)), then:

d_1 <= d_0 AND d_2 <= d_0.

Also, since A^2 and B^2 are odd (and so are sigma(A^2) and sigma(B^2)), then d_1 >= 3 AND d_2 >= 3.

Again, I ask the question: How can one use these facts, together with 8/5 < sigma(A^2)/A^2 = sigma(B^2)/B^2 < 2, to produce some numerical information about A^2 and B^2?