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Topic: n does not divide abc
Replies: 4   Last Post: Jul 15, 2014 2:18 AM

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robersi

Posts: 116
From: NYS
Registered: 8/18/09
n does not divide abc
Posted: Jul 15, 2014 12:28 AM
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Assume a^n + b^n = c^n where a,b,c are non zero integers and n is an odd prime.

and the gcd(a,b,c)=1

also assume


a^n-a=0 (mod n^2)
b^n-b=0 (mod n^2)
c^n-c=0 (mod n^2)

then there exists an x,y, and z such that

x+y-z=0
where 0<(x,y,z)<n
x^n+y^n=z^n mod(n^3)

also abc!=0 (mod n)

====================================================
let


a = x + nI_a
b = y + nI_b
c = z + nI_c


where 0<(x,y,z)<=n


a = -x' + n(I_a+1)
b = -y' + n(I_b+1)
c = -z + n(I_c+1)


-n<(-x',-y',-z')<=0 ;
0<=(x',y',z')<n
===================
x+x'=n
y+y'=n
z+z'=n
===================


also assume


a^n-a=0 (mod n^2)
b^n-b=0 (mod n^2)
c^n-c=0 (mod n^2)


therefore a+b-c=0 (mod n^2)


and then
because the gcd(a,b,c)=1

x+y-z=0 (mod n)
x'+y'-z'=0 (mod n)


but because


0<(x,y,z)<=n
and
0<=(x',y',z')<n


we have


x+y-z=0 or =n
x'+y'-z'=n or =0
==========================================


assume, for the time being, that


x+y-z=0


and


x'+y'-z'=n


then


nI_a + nI_b - nI_c = 0 (mod n^2)


a^n = x^n + (n^2)I_a(x^(n-1)) +n^3(I_a')
b^n = y^n + (n^2)I_b(y^(n-1)) +n^3(I_b')
c^n = z^n + (n^2)I_c(z^(n-1)) +n^3(I_c')


a^n = x^n + (n^2)I_a +n^3(I_a'')
b^n = y^n + (n^2)I_b +n^3(I_b'')
c^n = z^n + (n^2)I_c +n^3(I_c'')


a^n+b^n-c^n = 0 = x^n+y^n-z^n +(n^2)(I_a+I_b-I_c) +n^3(I')


x^n+y^n-z^n = -(n^2)(I_a+I_b-I_c)-n^3(I')


x^n+y^n=z^n mod(n^3)
---------------------------------------------------

OR we have (in the other case) simalarly


x'+y'-z'=0
where -n<(x',y',z')<=0
x'^n + y'^n - z'^n = 0 mod(n^3)
---------------------------------------------------------------------------------------------------------

Assume a, b, and c are non-zero integers and n is an odd prime.
a^n + b^n = c^n can be rewritten as A^n + B^n = C^n where a = A, b = B, and c = C.
a^n + (-c^n) = -b^n can be rewritten as A^n + B^n = C^n where a = A, -c = B, and -b = C.
b^n +(-c^n) = -a^n can be rewritten as A^n + B^n = C^n where b = B, -c = A, and -a = C.
So because the GCD(A,B,C) = 1 then it can can be said without a loss of generality that AB!= 0 mod n since
A^n + B^n = C^n can take anyone of these three forms.

------------------------------------------------------------------------------------------
therefore in either case if n|c then z=0 mod n

so either z=0 or z=n.

----------------------------------------------------------------------------------------




because gcd(a,b,c)=1 and we assumed in the first case (w.o.l.o.g.) that 0<(x,y,z)<=n

then only z can = n. ()

assume z = n

x^n + y^n = n^n mod (n^3)

x^n = -y^n mod (n^3)

therfore

x=-y mod (n^2)

contradiction

z!=n


abc!=0 mod n
---------------------------------------------------------------------------------------
recall
0<=(x',y',z')<n


OR we have (in the other case)


x'+y'-z'=0


x'^n + y'^n - z'^n = 0 mod(n^3)

----------------------------------------------------------------------------------------
--------------------------------------------------------------------------------------
So assume C=0 (mod n) ===> z' = 0 ====> since x'+y'=z' and -n<(x',y',z')<=0
and the gcd(a,b,c)=1
then z!=0 ==========> abc!=0 (mod n)


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