- #1
Victor Sorokine
- 70
- 0
Designations:
a_(i) - the i-th digit from the end in the number a in the prime base n> 2.
a_(i, j) - the number, comprised of the digits from i to j of the number a.
9 – designation of digit n-1,
8 – designation of digit n-2.
Proof of FLT
(01°) Let us assume the solution of the equation A^n+B^n-C^n = 0 in the integers exists.
Let us lead the number u=a+b-c to the form:
(02°) u = (n^k)(n^p-1)
with the aid of the multiplication of the equality of 01° by the sufficiently large number d^n (which, as is known, in the prime base n exists), as a result of which the equality of 01° is converted into the equality
(1°) a^n+b^n-c^n = 0, where c> a> b and a+b> c.
Let us break all digits of the number u into three intervals:
1) [p, q+1],
2) [q, r+1],
3) [r, 1],
where p is greatest rank of the number c, q is the greatest rank of the significant part of the number u, r is greatest rank of the zero end of the number u.
I.e. the number u = u_(p, q+1)n^q+u_(q, r+1)n^r+u_(r, 1) = /00… 00//99… 99//00… 00/.
Since the end u_(r, 1) (=0) does not change from the multiplication of the number u on d and the number of a^n+b^n-c^n (as we will see more lately) is considerably less than zero, then the endings a_(r, 1), b_(r, 1), c_(r, 1) can be rejected as infinitesimal quanitiies in comparison with the number of a^n+b^n-c^n.
And the now approximate values of the numbers a, b, c with their MAXIMUM value take the form:
(2°)
a=/99… 98//99… 99/+ 1,
b=/00… 00//99… 99/+ 1,
==============
c=/99… 99//99… 99/+ 1.
Or:
(3°)
a=n^{p-r}-n^{q-r},
b=n^{q-r},
========
c=n^{p-r}.
But with these values of the numbers a, b, c we have a equality:
(4°) a+b = c, and therefore
(5°) a^n+b^n-c^n<<0.
It is easy to see that an increase (decrease) in the numbers a, b, c by 1 does not can change inequality sign of 5° (with the sufficiently large number d from 02°).
It remains to add that the inequality of 5° remains true and for ANY positive numbers
(6°)
a=n^{pr} - n^{q- R} of+a',
b=n^{q- R}+b' ,
========
c=n^{pr}+b',
with the condition of a'+b'- c'=0 (and with the sufficiently large number d from 02°).
Thus, the expression 1° (or 01°) in FLT is inequality a^n+b^n-c^n<<0.
FLT is proven.
(On July 6, 2008. [Mezos])
a_(i) - the i-th digit from the end in the number a in the prime base n> 2.
a_(i, j) - the number, comprised of the digits from i to j of the number a.
9 – designation of digit n-1,
8 – designation of digit n-2.
Proof of FLT
(01°) Let us assume the solution of the equation A^n+B^n-C^n = 0 in the integers exists.
Let us lead the number u=a+b-c to the form:
(02°) u = (n^k)(n^p-1)
with the aid of the multiplication of the equality of 01° by the sufficiently large number d^n (which, as is known, in the prime base n exists), as a result of which the equality of 01° is converted into the equality
(1°) a^n+b^n-c^n = 0, where c> a> b and a+b> c.
Let us break all digits of the number u into three intervals:
1) [p, q+1],
2) [q, r+1],
3) [r, 1],
where p is greatest rank of the number c, q is the greatest rank of the significant part of the number u, r is greatest rank of the zero end of the number u.
I.e. the number u = u_(p, q+1)n^q+u_(q, r+1)n^r+u_(r, 1) = /00… 00//99… 99//00… 00/.
Since the end u_(r, 1) (=0) does not change from the multiplication of the number u on d and the number of a^n+b^n-c^n (as we will see more lately) is considerably less than zero, then the endings a_(r, 1), b_(r, 1), c_(r, 1) can be rejected as infinitesimal quanitiies in comparison with the number of a^n+b^n-c^n.
And the now approximate values of the numbers a, b, c with their MAXIMUM value take the form:
(2°)
a=/99… 98//99… 99/+ 1,
b=/00… 00//99… 99/+ 1,
==============
c=/99… 99//99… 99/+ 1.
Or:
(3°)
a=n^{p-r}-n^{q-r},
b=n^{q-r},
========
c=n^{p-r}.
But with these values of the numbers a, b, c we have a equality:
(4°) a+b = c, and therefore
(5°) a^n+b^n-c^n<<0.
It is easy to see that an increase (decrease) in the numbers a, b, c by 1 does not can change inequality sign of 5° (with the sufficiently large number d from 02°).
It remains to add that the inequality of 5° remains true and for ANY positive numbers
(6°)
a=n^{pr} - n^{q- R} of+a',
b=n^{q- R}+b' ,
========
c=n^{pr}+b',
with the condition of a'+b'- c'=0 (and with the sufficiently large number d from 02°).
Thus, the expression 1° (or 01°) in FLT is inequality a^n+b^n-c^n<<0.
FLT is proven.
(On July 6, 2008. [Mezos])