(cn/2 + bn/2) x (cn/2 – bn/2) = a1n x a2n = an (1)
Where a1n = cn/2 + bn/2 , a2n = cn/2 – bn/2 are real positive factors of the number an.
From (1) it follows: cn = (an1 + an2)/2)2, bn = (an1 - an2)/2)2 (2)
In accordance with the properties of the exponential function, for real positive numbers a1n , a2n and integer a ≠ 1, there are unique values of the exponents - <∞ q < p <+∞,, satisfying the equalities:
a n1 = a p, a2n = a q (3) , где a = a1 a2, n = p + q.
From (3) it follows a1p a1q = a1pa2p, a2p a2q = a1qa2q , or after reduction by numbers
a1p ≠ 0 , a2q ≠ 0 получим: a1 q = a2 p (4)
From (1), (2) and (3) it follows:
cn - bn = (ap +aq)/2)2 - (ap -aq)/2)2 = ap+q =an (5)
or, taking into account equalities (3) and (4):
(a1 p * a1 q +a1 q * a 1q)/2)2 - (a1 p * a1 q - a1 q * a 1q)/2)2 = a1 p * a1 2q * a2 q (6)
Let's take the common factor a1q out of brackets:
a1 2q * (a1 p +a2 q )/2)2 - a1 2q * (a1 p - a2q )/2)2 = a1 p*a1 2q * a2 q (7)
From (5) and (7) it follows that the numbers cn , bn and an contain a common factor a1 2q = a22 p ≠ 0, which contradicts the condition of their mutual simplicity, if a1 2q = a22 p ≠ 1. From a1 2q = a22 p = 1 it follows q = 0 , a2 = 1, that is, n = p , a = a1 *1, and equalities (5) and (7) take the form: cn - bn = (an +1 )/2)2 - (an -1 )/2)2 = an (8)
From (8) it follows that for odd a ≠ 1 the numbers cn and bn are also integers, and the identity always holds:
cn/2 - bn/2 = 1 (9)
that for simultaneously integer b, c and n is feasible only for n/2 ≤ 1, n = 1, 2, which is what needed to be proved.
The proof can be carried out in a slightly different way. All numbers of the equality an + bn = cn, where a ≠ 1, b and n are arbitrarily chosen natural numbers, c is a real positive number, through transformations (1)…(4) can be expressed as terms of identity (5).
Let us take the factor aq ≠ 0 out of brackets and divide all terms of identity (5) by it:
(ak +1)/2)2 - (ak - 1)/2)2 = ak (10) where k = p - q.
In accordance with the properties of the exponential function, arbitrarily chosen natural numbers a, b and n, for example, from equality (5), correspond to a single value k satisfying the condition:
bn = (ak - 1)/2)2 (11) then cn = bn + ak = (ak + 1)/2)2, or cn - bn = ak (12) where a, b and n are integers.
From (10), (11) and (12) it follows: : cn/ 2 - b n/2 = 1 (13) that is, the numbers b and c can be simultaneously integers only when n/2 ≤ 1, or n = 1, 2 . For n = 2, b and c are consecutive integers. Euclid also proved that any odd number can be expressed as the difference of the squares of two consecutive integers, which can be found using identity (10) for any integer k and odd a ≠ 1.
Note that equality (12) was obtained by dividing equality (5) by the factor a 2q, while the number bn in these equalities is the same, which 2q = 1, q = 0 , p = k = n, and the identity (10) takes the form of identity (8).
Note also that identities (8) and (10) are valid not only for integer values of a. By substituting any rational fraction for a and setting k = n = 2, you can find all Pythagorean numbers.
The above transformations of the Fermat equality over the set of natural numbers show that with the help of a finite number of arithmetic operations it is always reduced to identity (13), which proves the theorem.
2. Proof of Fermat's theorem using elementary algebra methods
Later, Bobrov A.V. published another proof of Fermat’s Theorem.
Fermat's theorem states that the equality an + bn = cn for natural a, b, n can only hold for integers n ≤ 2.
Consider the equality C – B = A, (1) where A, B, C are coprime natural numbers, that is, numbers that do not have common integer factors other than 1. In this case, two numbers are always odd. Let –A = an - an odd number, a ≠ 1 and n - natural numbers. For any real positive number, we can perform the operation of finding the arithmetic value of the root, that is, equality (1) can be written as:
(√C + √B) (√C –√B) = A1 A2 = A, (2) where A1 = √C+√B и A2 =√C-√B -
real positive factors of a number In accordance with the properties of the exponential function, for any of the real positive numbers A1 and A2 there are unique values of the numbers 0< q< p < ∞ satisfying the equalities A1 = a p A2 = a q, (3)
From equalities (2) and (3) it follows:
√C = (ap + aq)/2, √B = (ap - aq)/2, A = a p+q = a n (4)
Since p > q, always takes place p-q=k,, or аp= аk × аq, that is, the numbers A,B,C contain a common factor, which contradicts the condition of their mutual you just. This condition is fulfilled only when aq = 1, that is, when q= 0, p = n.
Then equalities (4) take the form:
√C = (an + 1)/2, √B = (an - 1)/2, A = a n (5)
whence follows
√C - √B = 1 (6)
that is, for coprime A,B,C, the numbers c =√C, b = √B are always two consecutive integers. Euclid also proved that every odd number is expressed as the difference of the squares of two consecutive integers, that is, equality (1) for natural coprime a, b, c can only be expressed as the equality c2 -b2 = a2. (7)
The validity of the above proof can be illustrated by the following example.
Let the numbers a,b,c in Fermat’s equality be coprime integers, and let n be even. Then the numbers cn/2, bn/2, their sum A1 = ap and difference A2 = aq are also integers, the exponent p>q.
Integers cn/2 = ( ak+q + aq)/2, bn/2 = (ak+q - aq)/2 are coprime if they do not contain common integer factors other than 1. This condition is satisfied only if the common integer factor is aq = 1, that is, q = 0, p = n.
Then the difference cn/2 - bn/2 = 1, which for simultaneously integer b,c,n can only occur when n/2 ≤ 1, that is, when n = 1, 2.
2. Second proof 2001-2004
In 2001-2004, V. A. Gorbunov published a simple proof of Fermat’s last theorem at MSTU.
Proof of Fermat's Last Theorem
Fermat's theorem states that there are no integer solutions to the equation
xn + yn = zn (1) for any natural number n > 2 (xyz ≠0).
Proof
By introducing the transformations u = x/z, v = y/z, (2) equation (1) is reduced to the form un + v n = 1. (3)
It is obvious that the variables u and v satisfy the inequalities 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1. (4)
Let us denote the sum un + vn by Sn , and call the graph of the equation Sn = 1 the Fermat curve of the n-order. Figure 1 shows a square containing: a unit circle S2 =1 and a Fermat curve Sn =1 (n ≥ 3). Due to the symmetry of equation (3) with respect to the variables u and v, in what follows we will consider only the upper part of the square with respect to the secondary diagonal (matrix terminology). For any natural number z (z > 1), we apply a uniform grid on the unit square with a rational step h = z/2, Figure 1. The larger z, the smaller h, that is, the denser the grid in the unit square. The statement of Fermat's theorem is equivalent to the fact that whatever the number z (and, consequently, h), Fermat curves do not pass through the nodal (rational) points in the unit square, which is what remains to be proved.
Fig.1. Unit circle S2 =1 and Fermat curve Sn = 1 (n =3,4,…) in a unit square covered with a uniform grid with a step h =1/z.
It is known that there is an infinite number of rational points on the unit circle. If, when dividing a unit square for a natural number z0, the rational point of the unit circle P0 (x0/z0, y0/z0) turns out to be a nodal point on the mesh, then we will call this point Pythagorean, and the ray passing through it (coming from the origin) will be called Pythagorean beam ƛ0.
If the Pythagorean number z0 is prime, then at the top of the unit square on the unit circle there will be one Pythagorean point on the grid for z0.
1. Any point in the unit square M0 (x0/z1, y0/z2) with rational coordinates can become a node on the grid for a natural number z = z1 * z2, if NOD (z1 * z2) = 1= 1. Thus, in what follows about a rational point, we will assume that the denominators of the coordinates of the point are the same.
For composite numbers of the form z = k * p1* p2 *...*pl,, where p1 p2 ,..., pl are prime numbers of the form 4n+1, and k is a natural number whose factorization does not contain Pythagorean numbers, then in the upper part of the unit square there will be l2 Pythagorean points on the unit circle for a grid with a step h = 1/ z.
Let us denote by ƛ the ray forming an angle ƛ with the Ou axis. In addition to the designation, we will give a numerical value to the parameter ƛ/ƛ = tgȹ. Then the coordinates of the points on this ray are related by the relation v = ƛ*u. Substituting instead of v into equation (3) ƛ*u, we obtain the coordinates of the point Фn lying at the intersection of the ray ƛ with the Fermat curve Sn =1:
un = 1/ n√(1+ ƛ n), vn = ƛ / n√(1+ ƛ n) (5)
In what follows, we will be interested only in rays passing through nodal (rational) grid points in a unit square. On such rays the parameter ƛ takes rational values. We will distinguish between rays ƛ 0 passing through rational points on the unit circle and rays ƛ* passing through rational points in the unit square and not lying (points) on the unit circle.
Let us prove a lemma that will be useful to us later.
