Re: (2^n-1)(3^n-1) never a square ?

Am 24.01.2008 13:28 schrieb atul:
A not so rigorous proof.

Say (2^n - 1) (3^n - 1) = m^2 = p1^2a1 . p2^2a2 . ... . pk^2ak

Each pi must go in 2^n -1 or 3^n -1, either way if pi's are not 2 or
3, then:
=> _PI_ ( pi^2ai - 1) | n (Fermat's Little Theorem or Gauss's
reciprocals

LHS is comparable to m^2 which is >> n

Q.E.D.

Don't waste time on this, I am not correct.

Perhaps the following is of a bit help here.
From the cyclicity of perimefactorization of
cyclotomic expressions like a^n -1 we may derive
some conditions on m in (4^m-1)*(9^m-1)
(We know already, that in (2^n-1)(3^n-1) n must be
even and n=2m)

I introduce two symbold for conciseness of notation

Let a:p = 1 if p divides a , and a:p=0 if not. If a=0, then a:p=0.

Let {a,p} the p-adic-valuation of a , where p is prime,
so if a=5*2^4 then {a,2}=4 {a,5}=1 and {a,3}=0

Now let us allow to write
a:2 * {a,3} = x
meaning if 2 divides a, then x=valuation(a,3) , so
18 : 2 {18,3} = 1 * 2 = 2
9 : 2 { 9,3} = 0 * 2 = 0

Then we can compactly denote combined conditions on m to
satisfy
T = T4*T9= (4^m - 1)*(9^m - 1) = prod{p_k=primes}( p_k^(2*i_k))
where i_k is a factor specific to p_k

Example:

Prime Expression for exponent of prime
2 {T4,2} = 0
{T9,2} = 3 + {m,2}
{T,2} = {T4,2} + {T9,2} = 3 + {m,2}
This means, m must contain an odd power of 2, such that {T,2} is even.


Prime Expression for exponent of prime
3 {T4,3} = 1 + {m,3}
{T9,3} = 0
{T,3} = 1 + {m,3}
This means, m must contain an odd power of 3, such that {T,3} is even.


From this two examples I already know, that m must contain an
odd power of 2 as well as an odd power of 3 to have the primes 2 and 3
as squares in the value of T.

This means: m = x * 2*4^k2 * 3*9^k3
where {x,2}={x,3}=0 and k2 and k3 are natural numbers>=0

Prime Expression for exponent of prime
5 {T4,5} = m:2 * {m,5}
{T9,5} = m:2 * {m,5}
{T,5} = m:2 * 2{m,5}
means: if m is even, then {T,5} = 2 k5 is even with k5>=0
Since m must be even as already known, we have always the
primefactor 5 with an even exponent in T.
So the primefactor 5 does not give more restrictions on m.

Prime Expression for exponent of prime
7 {T4,7} = m:3 * {m,7}
{T9,7} = m:3 * {m,7}
{T,7} = m:3 * 2{m,7}
analoguously as before replacing cycle-index 2 by cycle-index 3 in m:2 resp m:3

Prime Expression for exponent of prime
11 {T4,7} = m:5 * {m,11}
{T9,7} = m:5 * (1+{m,11} )
{T,7} = m:5 * (1+2{m,11})
Here we find, that the valuation (T,11) is always odd; so it must
be m:5 = 0 to prevent the occurence of an odd number of primefactor 11.

So m must have the form (with a natural x not containing 2,3,5 as prime-factors)

m = 2*4^k2 * 3*9^k3 * 5^0 * x

But if m contains powers of 2, say m:4 =1 , then T contains also
several more primefactors with the same cycle length, for instance
p=17. And if T contains p=17 then it must contain it to the
power of 2. Well, this occurs with most primes p_k for T4 and T9
parallel, to T4*T9 (using the restricted m) having p_k to
an even power is not much restrictive. Except, for example, p_k=11,
which is a wieferich to base 3 but not to the base 2, definitely
restricts m to be non-divisible by 5.

I checked a handful of more primefactors but without finding
more decisive general exclusion-arguments based on the last
consideration, but perhaps one may find a certain combined
restriction, which shows the impossibility of a square T
in general, at least for m > certain lower bound.

Possibly there occurs another option. Since m contains factors
of 2 raised to even powers, T4 and T9 can be factored into many
disjunct factors. Let r = m/ 4^k2 then

T4 = (4^r)^(4^k2) - 1 = (2^r - 1)
*(2^r + 1)
*((2^r)^2 + 1)
*((2^r)^4 + 1)
*((2^r)^8 + 1)
*...
*((2^r)^(2^k2) + 1)

and T9 has similarly many factors. Also because
r is composite with powers of 3, we may formulate another
sequence of factors, where the sheer number outperforms
the possibility that they all can occur as sqaures.
But I didn't consider this in more detail.

Gottfried

--
---

Gottfried Helms, Kassel
.

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