Page 145 - ASBIRES-2017_Preceedings

NEW SIMPLE PROOFS OF FERMAT’S LAST THEOREM FOR N=3

area of a Pythagorean triangle cannot be a +12ℎ < 0. This is impossible since ℎ
square of an integer have been proved using positive and we conclude that there are no
two simple independent algorithms. A non trivial integer triples satisfying (1).
simple and simple proof of Fermat’s last
theorem for = 3 is also discussed to point 3 ANOTHER PROOF OF FERMAT’S
out that the method of infinite descent may LAST THEOREM FOR = 3
be a tailor-made method by Fermat for the According to the equation (3),
proof of above two theorems.

3
2
+ 3 + 3 2 − 1 = 0 since
2 NEW PROOF OF FERMAT’S LAST 2
THEOREM FOR =3 ℎ =

Fermat’s last theorem for = 3 can In other words, + 3 +
3 2
2
be stated as that there are no non-trivial 3 − = 0, is an equation for . We
2
2
integers satisfying the equation (1) in the look for integer solutions for in (5) and we
following. Let us assume that there are obtain
, , > 0 integers which satisfy (1) in the
2
2 2
2
following. [ + 3 + 3 ] = (4)
3
3
3
= + , (, ) = 1 (1) Without loss of generality we can
2
assume that (, 3) = 1. Then = , = 1
The equation (1) can be written as are possible integer solutions for . If =

1, we have − = 1 and it gives = +
3
3
= ℎ + 1 (2)

. From which we obtain
3
3
Where = 3 , ℎ = 3 (> 0 ) 2 2
3 3 3 + 3 = 3( + ) = 0
From the equation (2), we get The equation (6) implies that one
of(, , ) is zero since + is also factor
2
2
3
3
− ℎ = ( − ℎ)( + ℎ + ℎ ) = 1. 3 2
of which follows from (1). If = ,
2
1
2
Now, let − ℎ = since + 3ℎ = , then = + , from which we obtain
3 6
3
2
2
2 4
= 3 + 3 + which does
> 0. not hold since and , are positive
6
+ 3ℎ − 1 = 0 and since − ℎ = integers and as a result of this right – hand
3
, + 3( + ℎ)ℎ − 1 = 0 and therefore value is greater than the left – hand value.
3

Assume that = ,(, ) = 1and ≠ 1.

3
2
2
+ 3ℎ + 3ℎ − 1 = 0 (3)
Then from (5), we get
It should be noted that Cardon and [ + 3 + 3 ] = , = 1,
2 2
2 2
3 2
Tartagalia method fails in obtaining the =
2
solution for in (3) (Archbold, 1961).
Where is a factor of . Similarly
2
2
Now, it is clear that 3ℎ + 3ℎ − 1 < 0 = 1 or = where is a factor of and
2
since is positive. Writing this as (, ) = 1 and since ≠ 1 it is possible that
2

= − , we
2
= and then since =
−3ℎ − 3ℎ + 1 > 0 (4)
2
2
2

1
must have − = 2 which is not an
2
We get the quadratic −3ℎ − 2
2
3ℎ + 1 is positive and hence its integer. Therefore cannot be a rational

discriminant should be negative. This is the number of the form where ≠ 1. Hence,
necessary condition for the above quadratic we conclude that there are no non trivial
4
to be of the same sign. In other words 9ℎ integers , , satisfying the equation (1).
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