Thread View: sci.logic
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Started by hporopudas
Thu, 29 Oct 1992 07:31
Inequalities for Fermat's last theorem
Author: hporopudas
Date: Thu, 29 Oct 1992 07:31
Date: Thu, 29 Oct 1992 07:31
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In article <1992Oct28.161726.1@tnclus.tele.nokia.fi>,
hporopudas@tnclus.tele.nokia.fi writes:
>
> Hello !
>
> Does any of you know inequalities related to z in Fermat's last theorem.?
> n n n
> ( x + y = z ; x,y,z,n positive integers, x < y, n > 2 )
>
> ---Hannu.
>
I have discovered some years ago following inequalities for x, y, z and n
(int(v)=integer part of v, log is natural logarithm):
-----------------------------------------------------------------------
! log2 !
! 3 < n < int( ---------- ) ! (1)
! = = z !
! log(-----) !
! z - 1 !
-----------------------------------------------------------------------
! z z !
! 1 < x < int( ------- ) < ------- ! (2)
! = = 1/n = 1/n !
! 2 2 !
-----------------------------------------------------------------------
! z z !
! ---- < int( ----- ) + 1 < y < (z - 1) ! (3)
! 1/n 1/n = = !
! 2 2 !
-----------------------------------------------------------------------
! n n 1/n n z n 1/n !
! z-int(z -(z -(z-1) ) ) < x < z-int(z -(z -(int(------)+1) ) ) ! (4)
! = = 1/n !
! 2 !
------------------------------------------------------------------------
Also the following inequalities holds:
n
n n n z
z - (z-1) < x < --- (5)
2
2
x xy
-------- < -------- < z (6)
n 1/n n 1/n
(x -1) (x -1)
1/n
2
------- < z (7)
1/n
2 -1
n n
1 -(x - 1) 1 x - 1 2 1 1/2
---- > --------- + ------- ( (--------) + --- ) (8)
n n n 2 n n n n 2
z (y - x ) (y - x ) y - x
Fermat's equation has also the following alternative series form:
----------------------------------------------------------
! n 2n 3n !
! z - x y -n (n-1) y -2n (n-1)(2n-1) y -3n !
! ----- = --- z + --------- z + -------------- z +...! (9)
! z n 2 3 !
! 2! n 3! n !
----------------------------------------------------------
! n !
! y k1 !
! --- = constant z + .... , 0 < constant < (z-1) ! (10)
! n = = !
! n !
! y !
! where k1 = int( log ( ---- )), (base-z logarithm) !
! z n !
! !
----------------------------------------------------------
---Hannu
Re: Inequalities for Fermat's last theorem
Author: hporopudas
Date: Fri, 30 Oct 1992 10:19
Date: Fri, 30 Oct 1992 10:19
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In article <1992Oct29.093106.1@tnclus.tele.nokia.fi>,
hporopudas@tnclus.tele.nokia.fi writes:
>
> In article <1992Oct28.161726.1@tnclus.tele.nokia.fi>,
> hporopudas@tnclus.tele.nokia.fi writes:
>>
>> Hello !
>>
>> Does any of you know inequalities related to z in Fermat's last theorem.?
>> n n n
>> ( x + y = z ; x,y,z,n positive integers, x < y, n > 2 )
>>
>> ---Hannu.
>>
>
> I have discovered some years ago following inequalities for x, y, z and n
> (int(v)=integer part of v, log is natural logarithm):
>
> -----------------------------------------------------------------------
> ! log2 !
> ! 3 < n < int( ---------- ) ! (1)
> ! = = z !
> ! log(-----) !
> ! z - 1 !
> -----------------------------------------------------------------------
> ! z z !
> ! 1 < x < int( ------- ) < ------- ! (2)
> ! = = 1/n = 1/n !
> ! 2 2 !
> -----------------------------------------------------------------------
> ! z z !
> ! ---- < int( ----- ) + 1 < y < (z - 1) ! (3)
> ! 1/n 1/n = = !
> ! 2 2 !
> -----------------------------------------------------------------------
> ! n n 1/n n z n 1/n !
> ! z-int(z -(z -(z-1) ) ) < x < z-int(z -(z -(int(------)+1) ) ) ! (4)
> ! = = 1/n !
> ! 2 !
> ------------------------------------------------------------------------
>
>
> Also the following inequalities holds:
>
> n
> n n n z
> z - (z-1) < x < --- (5)
> 2
> 2
> x xy
> -------- < -------- < z (6)
> n 1/n n 1/n
> (x -1) (x -1)
>
> 1/n
> 2
> ------- < z (7)
> 1/n
> 2 -1
> n n
> 1 -(x - 1) 1 x - 1 2 1 1/2
> ---- > --------- + ------- ( (--------) + --- ) (8)
> n n n 2 n n n n 2
> z (y - x ) (y - x ) y - x
>
>
>
> Fermat's equation has also the following alternative series form:
>
> ----------------------------------------------------------
> ! n 2n 3n !
> ! z - x y -n (n-1) y -2n (n-1)(2n-1) y -3n !
> ! ----- = --- z + --------- z + -------------- z +...! (9)
> ! z n 2 3 !
> ! 2! n 3! n !
> ----------------------------------------------------------
> ! n !
> ! y k1 !
> ! --- = constant z + .... , 0 < constant < (z-1) ! (10)
> ! n = = !
> ! n !
> ! y !
> ! where k1 = int( log ( ---- )), (base-z logarithm) !
> ! z n !
> ! !
> ----------------------------------------------------------
>
>
>
> ---Hannu
As a matter of fact this constant > 0. (k1 = k_1)
Similarly
2n
(n-1) y k_2
--------- = constant z + ... , 0< constant < (z-1)
2 2 2 =
2! n
where
2n
(n-1) y
k_2 = int ( log ( --------- ))
z 2
2! n
... ... ... ... ...
in
(n-1)(2n-1)... ((i-1)n -1) y k_i
----------------------------- = constant z + ... ,
i i
i! n
0 < constant < (z-1) and
i =
in
(n-1)(2n-1)... ((i-1)n -1) y
k_i = int( log ( ----------------------------- ))
z i
i! n
(Idea is to try to prove that (9)'s right side is unperiodic number
in the base system z.).
---Hannu.
P.S. My employer denied me to write anything in sci.physics, sci.astro,
sci.math, sci.philosophy.tech and talk.religion.misc because some from
sci.physics or sci.math contacted to my employer due to my posting
'Structure of Time' (which article was spoked in the name of God by my
5 years old (28.4.1992) daughter Hanna-Maria).
Re: Inequalities for Fermat's last theorem
Author: hporopudas
Date: Mon, 02 Nov 1992 07:07
Date: Mon, 02 Nov 1992 07:07
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In article <1992Oct29.093106.1@tnclus.tele.nokia.fi>,
hporopudas@tnclus.tele.nokia.fi writes:
One correction to (8):
>
> n n
> 1 -(x - 1) 1 x - 1 2 1 1/2
> ---- > --------- + ------- ( (--------) + --- ) (8)
> n n n 2 n n n n 2
> z (y - x ) (y - x ) y - x
>
(8) should be:
n n
1 -2(x - 1) 2 x - 1 2 1 1/2
---- > ---------- + ------- ( (--------) + --- ) (8)
n n n 2 n n n n 4
z (y - x ) (y - x ) y - x
This has also the form:
n n n 2 n n 2 (1/2)
z < 2 ((x -1) + ((x -1) + (1/4)( y - x ) ) ) (8')
Could any of you say where the error in the following reasonig ?:
-----------------------------------------
Combining (8') and (7) we have also
n 2 n n 2 (1/2) 1 n n
((x -1) + (1/4)( y - x ) ) > ( ------ ) - (x -1)
1/n
2 -1
From
n n 1 2n n 1 n (1/2)
y - x > 2( ( ------- ) - 2( x - 1) ( ----- ) )
1/n 1/n
2 -1 2 -1
This also means
n 1 n
x < (1/2)( ------ ) + 1
= 1/n
2 -1
This last inequality and the result of prof. Inkeri
(finnish mathematician), which states that
3n-4
x > n ; for n > 3
=
(limitations?)
We have achieved the following impossible inequality
3n-4 1 n (1/n)
n < x < ((1/2)( ------ ) + 1)
= 1/n
2 -1
because left side of the the inequality grows always faster than right
side of the inequality, when n is greater than 2.
---Hannu.
Re: Inequalities for Fermat's last theorem
Author: hporopudas
Date: Tue, 03 Nov 1992 10:33
Date: Tue, 03 Nov 1992 10:33
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In article <1992Nov2.090701.1@tnclus.tele.nokia.fi>,
hporopudas@tnclus.tele.nokia.fi writes:
...deleted part...
>
>
> Could any of you say where the error in the following reasonig ?:
> -----------------------------------------
>
> Combining (8') and (7) we have also
>
> n 2 n n 2 (1/2) 1 n n
> ((x -1) + (1/4)( y - x ) ) > ( ------ ) - (x -1)
> 1/n
> 2 -1
> From
> n n 1 2n n 1 n (1/2)
> y - x > 2( ( ------- ) - 2( x - 1) ( ----- ) )
> 1/n 1/n
> 2 -1 2 -1
>
This last inequality is questinable ,because by suitable rearragements of
terms you get also
n n 1 2n n 1 n (1/2)
y - x < 2( ( ------- ) - 2( x - 1) ( ----- ) )
1/n 1/n
2 -1 2 -1
>
> This last inequality and the result of prof. Inkeri
> (finnish mathematician), which states that
>
> 3n-4
> x > n ; for n > 3
> =
> (limitations?)
>
Corrected form of this K. Inkeri's result is
3n-4
x > (1/2) n ; for n is prime exponet > 2
(0 < x < y < z, x,y,z are relatively-prime integers)
Two cases of Fermat's last theorem:
----------------------------------
3
2n + n n
If (xyz,n)=1, (Case I), then x > ( -------- ) , n is prime exponent > 2.
log(3n)
(Case II is when n divides z, n is prime exponent > 2).
Reference:
Inkeri, K, 1953.
Absch{tzungen fur eventuelle L|sungen der Gleichung
im Fermatschen Problem.
Ann. Univ. Turku, Ser. A, vol. 16, no. 1, pages 3-9.
> We have achieved the following impossible inequality
>
> 3n-4 1 n (1/n)
> n < x < ((1/2)( ------ ) + 1)
> = 1/n
> 2 -1
>
> because left side of the the inequality grows always faster than right
> side of the inequality, when n is greater than 2.
>
I must reinvestigate what is followed from combining (8') and (7).
Above impossible inequality is under adjustment at the moment. It
is possibly wrong.
---Hannu.
Re: Inequalities for Fermat's last theorem
Author: hporopudas
Date: Wed, 04 Nov 1992 08:44
Date: Wed, 04 Nov 1992 08:44
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In article <1992Nov3.123313.1@tnclus.tele.nokia.fi>,
hporopudas@tnclus.tele.nokia.fi writes:
..all deleted from above message...
So we have:
Following inequality holds:
n n n 2
y - x < (x -1) - 1
From which we have the inequality:
2
z < x
K. Inkeri's result:
------------------
3n-4
x > (1/2) n ; for n is prime exponet > 2
(0 < x < y < z, x,y,z are relatively-prime integers)
(It can also be assumed without loss of generality that
x and y are relatively prime to p).
Two cases of Fermat's last theorem:
----------------------------------
3
2n + n n
If (xyz,n)=1, (Case I), then x > ( -------- ) , n is prime exponent > 2.
log(3n)
(Case II is when n divides z, n is prime exponent > 2).
Reference:
Inkeri, K, 1953.
Absch{tzungen fur eventuelle L|sungen der Gleichung
im Fermatschen Problem.
Ann. Univ. Turku, Ser. A, vol. 16, no. 1, pages 3-9.
I remove this discussion to sci.math.research, because I received
here in sci.logic none valuable comments.
---Hannu.
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