Jun 12, 2018 - The French mathematician Pierre de Fermat (1601-1665), conjectured that ... Fermat, Euler and Dirichlet solved the equation for n=3, 4 and 5.
Fermat Last Theorem The French mathematician Pierre de Fermat (1601-1665), conjectured that the equation x^n +y^n =z^n has no solution in positive integers x, y and z if n is a positive integer >= 3. He wrote in the margin of his personal copy of Brachet’s translation of Diophantus’ Arithmetica:”I have discovered a truly marvellous demonstration of this proposition that this margin is too narrow to contain”. Many researchers believe that Fermat does not find a demonstration of his proposition but some others think there is a proof and Fermat’s claim seems right. The search of a solution of equation x^n + y^n =z^n are splitted in two directions. The first one is oriented to search a solution for a specific value of the exponent n and the second is more general, oriented to find a solution for any value of the exponent n. - Babylonian studied the equation x^2+y^2=z^2 and found the solution (3,4,5). - Arabic mathematician Al-Khazin studied the equation x^3+y^3=z^3 in the X century and his work mentioned in a philosophic book by Avicenne in the XI century. Fermat, Euler and Dirichlet solved the equation for n=3, 4 and 5. - In 1753, Leonhard Euler presented a proof for x^3+y^3=z^3 - Fermat found a proof of x^4 +y^4 =z^4 using his famous “infinite descente”. This method combines proof by contradiction and proof by induction. - Dirichlet solved the equation x^5+y^5=z^5 - Sophie Germain generalized the result of Dirichlet for prime p if 2p+1 is prime.. Let p prime, x^p +y^p =z^p has no solution in positive integers if 2p+1 is prime. - In XIX century E.Kummer continued the work of Gauss and innovated by using numbers of cyclotomic field and introduced the concept of “prime factor ideal”. -Andrew Wiles, a professor at Princeton University, provided an indirect proof of Fermat’s Conjecture in two articles published in the May 1995 issue of Annals of Mathematics. In this paper, I would like to suggest a simple proof using mathematic tools of the Fermat‘s era and valid for whatever value n > 2.
x^n +y^n =z^n with n>=3 and 1 < x < y < z (or 1 3^n 5/ 3^n+2^n > 3^n 6/ 3^n+3^n > 3^n
The case z = 3 does not need any proof, because you can see that regardless of what you choose for x, y, or n, it is true that x^n + y^n ≠ 3^n. Just look at a few cases for x and y, for any n. The case z=2 implies the case z=3
x^n+y^n /=2^n 1+1< 2^n 1+2^n > 2^n 2^n+2^n > 2^n
x^n+y^n /=3^n implies … implies
1+1< 3^n 1+2^n < 3^n 2^n+2^n < 3^n
From the inequalities of z=2 (left column) we infer the inequalities of the right column which prove that the sums (x^n+y^n) are not equal to z^n=3^n in the case z=3. From the inequalities of SupG2 we infer the two last inequalities (2 and 3) of InfG3. SupG2: 1/ From 1+2^n > 2^n we infer 1 +2^n < 3^n 2/ From 2^n+2^n > 2^n we infer 2^n+2^n < 3^n
The inequality 3 from InfG3 is related to the case x=y in the Fermat equation (2*x^n=z^n) which obviously has no solutions in positive integers because 2=(3/2)^n is impossible. In the 6 inequalities, the sum ( x^n+y^n) is not equal to z^n= 3^n.
All the elements of the symmetrical matrix (3,3) with the form (x^n+y^n) are not equal to 3^n. Approach based on overlapping of the triangular matrixes The elements of matrix (2,2) are contained in the matrix (3,3). These elements are strictly smaller than 2^n then they are also strictly smaller than 3^n. These elements are: 1+ (2)^n, 2^n+(2)^n, 1+ (2)^n < 3^n and (2)^n + (2)^n=2(2)^n, we know that 2(2)^n = 3^n has no solutions in integers x, y , n>2 because 2 =(3/2)^n is impossible. There is no fraction whose power is 2. These elements vary from 1+ (2)^n 2.
4/
x^n +y^n=4^n with z=4 and 1 < x < y < 4 (or 12 x^n
y^n 1 2^n 3^n 4^n z^n
1
2^n
3^n
4^n
1+1
2^n +1 2^n+2^n
3^n +1 3^n +2^n 3^n +3^n
4^n+1 4^n+2^n 4^n+3^n 4^n+4^n
4^n
4^n
4^n
4^n
Approach based on the values of S=x^n+y^n compared to p^n There are 4*5/2= 10 different values. We can split the values of the sum (x^n+y^n) in two groups: InfG4 strictly smaller than 4^n and SupG4 strictly greater than 4^n . InfG4: 1/ 1+1 < 4^n 2/ 1+2^n < 4^n 3/ 1+ 3^n < 4^n 4/ 2^n +2^n < 4^n 5/ 3^n+2^n < 4^n 6/ 3^n+3^n < 4^n SupG4 7/ 1+4^n > 4^n 8/ 2^n +4^n > 4^n 9/ 3^n+4^n > 4^n 10/ 4^n+4^n > 4^n The case z=2 implies the case z=3
x^n+y^n /= 3^n 1+1< 3^n 1+2^n3^n 3^n+3^n >3^n
x^n+y^n /= 4^n implies …
implies
1+1< 4^n 1+2^n 3^n we infer 2^n+3^n < 4^n
The inequality 3 from InfG3 is related to the case x=y in the Fermat equation (2*x^n=z^n) which obviously has no solutions in positive integers.
In the 10 inequalities, the sum ( x^n+y^n) is not equal to z^n= 4^n.There is no equality.
All the elements of the symmetrical matrix (4,4) with the form (x^n+y^n) are not equal to 4^n. Approach based on overlapping of triangular matrixes The elements of matrix (3,3) are contained in the matrix (4,4). These elements are strictly smaller than 3^n then they are also strictly smaller than 4^n. We have to examine the new elements of the column (p-1=3) which is added to the matrix (3,3) when z is increased from 3 to 4. These elements are: 1+ (3)^n, 2^n+(3)^n, 3^n +(3)^n, 1+ (3)^n < 4^n and (3)^n + (3)^n=2(3)^n, we know that 2(3)^n = 4^n has no solutions in integers x, y , n>2 because 2 =(4/3)^n is impossible. These elements vary from 1+ (3)^n < 4^n to (3)^n + (3)^n 5^n In general (p-1)^n + (p-1)^n > p^n whatever p > 4 InfG4: 1/ 1+1 < 5^n 2/ 1+2^n < 5^n 3/ 1+ 3^n < 5^n 4/ 1+4^n < 5^n 5/ 2^n +2^n < 5^n 6/ 3^n+2^n < 5^n 7/ 4^n+2^n 5^n 11/ 1+ 5^n > 5^n 12/ 2^n +5^n > 5^n 13/ 3^n+5^n > 5^n 14/ 4^n+5^n > 5^n 15/ 5^n+5^n > 5^n The case z=4 implies the case z=5
x^n+y^n /= 4^n 1+1< 4^n 1+2^n < 4^n 1+3^n < 4^n 2^n+2^n < 4^n 1+4^n > 4^n 3^n+2^n >4^n 4^n+2^n >4^n 3^n+3^n >4^n 3^n+4^n > 4^n 4^n+4^n > 4^n
x^n +y^n /= 5^n implies …
implies
1+1< 5^n 1+2^n < 5^n 1+3^n < 5^n 2^n+2^n < 5^n 1+4^n < 5^n 3^n+2^n < 5^n 4^n+2^n < 5^n 3^n+3^n < 5^n 3^n+4^n < 5^n 4^n+4^n > 5^n
From the inequalities of the left column we infer the inequalities of the right column which prove that the sum (x^n+y^n) in the case z=5 are not equal to z^n=5^n. From the inequalities of SupG4 we infer the two last inequalities (3 and 5) of InfG5. SupG3: 1/ From 1+3^n > 4^n we infer 1 +3^n < 5^n from Binomial formula x^n+y^n < (x+y)^n 2/ From 2^n+3^n > 4^n we infer 2^n+3^n < 5^n
The inequality 3 from InfG4 is related to the case x=y in the Fermat equation (2*x^n=z^n) which obviously has no solutions in positive integers. In the 15 inequalities, the sum ( x^n+y^n) is not equal to z^n= 5^n.There is no equality. We can claim that x^n+y^n=4^n has no solutions in integers x, y and z=5 whatever n >2.
All the elements of the symmetrical matrix (5,5) with the form (x^n+y^n) are not equal to 5^n. Approach based on overlapping of the triangular matrixes The elements of matrix (3,3) are contained in the matrix (4,4). These elements are strictly smaller than 3^n then they are also strictly smaller than 4^n. We have to examine the new elements of the column (p-1=4) which is added to the matrix (3,3) when z is increased from 4 to 5. These elements are: 1+ (4)^n, 2^n+(4)^n, 3^n +(4)^n, 4^n + (4)^n, 1+ (4)^n < 5^n and 2^n +4^n < 5^n and 3^n +4^n < 5^n (4)^n + (4)^n=2(4)^n, we know that 2(4)^n = 5^n has no solutions in integers x, y , n>2 because 2 =(5/4)^n is impossible. These elements vary from 1+ (4)^n < 5^n to (4)^n + (4)^n /= 5^n .
The sum (x^n + (4)^n) is positive and increasing. In the case z=5 the elements of the column (p-1=4) are less than 5^n but the element on the diagonal of matrix (4,4) is greater than 5^n ( 4^n +4^n > 5^n). At one value of x (let x=k+1 < (4)) the function (x^n + (4)^n) switches from less than p^n to greater than p^n. (k+1)^n + (4)^n = (k^n +k^(n-1)+……1) + (4)^n=( k^n + (4)^n )+R with R>0 We know for k=3 3^n + (4)^n < 5^n then (3+1)^n + (4)^n < 5^n +R or (4)^n + (4)^n > 5^n All the elements of the column (p-1=4) are not equal to 5^n We can claim that x^n+y^n=5^n has no solutions in integers x, y and z=5 whatever n >2.
We have presented the examples z=3 , z=4 and z=5 in order to clarify the role of the implication between two consecutive cases (two consecutive values of z) and the overlapping of the triangular matrixes. In the induction step we will show that the case z=(p-1) implies the case z=p and the overlapping of the consecutive matrixes .
B. Inductive step Now assume the Fermat’s conjecture is true until z=(p-1). We know that every equation x^n + y^n =z^n from z=1 and z=2, z=3 until z=(p-1) has no solutions in integers x and y whatever n value > 2. We want to prove that FLT is true for z=p.
5/ z =p-1 x^n y^n 1 2^n 3^n 4^n
x^n + y^n =(p-1)^n
x< (p-1) and y < (p-1) and n >2
1
2^n
3^n
4^n
(p-1)^n
p^n
1+1
2^n+1 2^n+2^n
3^n+1 3^n+2^n 3^n+3^n
4^n+1 4^n+2^n 4^n+3^n 4^n+4^n
(p-1)^n+1 (p-1)^n+ 2^n (p-1)^n+ 3^n (p-1)^n + 4^n
p^n+1 p^n+2^n p^n+3^n p^n+4^n
(p-1)^n+(p-2)^n (p-1)^n+(p-1)^n
p^n+(p2)^n p^n+(p1)^n p^n+ p^n
z^n
z^n
(p-2)^n+(p-2)^n (p-1)^n p^n z^n
z^n
z^n
Z^n
z^n
Let us consider the next number p after (p-1), we have to show that x^n +y^n =p^n verifies Fermat’s conjecture. x^n +y^n =(p-1)^n has no solutions in integers x and y whatever n value > 2.
Approach based on the values of S=x^n+y^n compared to p^n There are p(p-1)/2 inequalities. We can split these inequalities of z= (p-1) in two groups InfGpm1 and SupGpm1.
InfGpm1 1+1
