Five proofs of the theorem of Fermat on five Author pages
Degtyaryov Alexander Vasilyevich
of BRD, Friedrich - Richter Str. 51,13125, Berlin, MO: 01639659078, Mail: degtjarewalex@mail. ru
Elementary proof of the theorem of Fermat
1. We use the small theorem of Fermat for the proof of the Great theorem Fermat. Method of cloning of the equations.
(if in hr x - natural number, and r- the prime number, then hr - x shares on r
without the rest)
we Will consider Fermat`s equation in the third degree h3 + u3 =z3. (1)
we Will transform it as follows: h3 - x + u3 - at - (z3- z) + x + at - z = 0,
h3 - x; u3 - at; (z3- z) share on 3 without the rest, therefore, x + at - z too of
shares on 3 without the rest, (at the same time we assume that (1) is carried out),
i.e. x + at - z = 3n1 (2)
where n1 - natural number. we Will present
(1) in the following form: h3 - h2 + u3 - u2 - (z3- z2) + h2 + u2 - z2 =0
or x (h2 - x) + at (u2 - at) - z (z2 - z) + x2 + u2 - z2 =0 (3)
Members of the equation (3) x (h2 - x); at (u2 - at); z (z2 - z) share on 2 without the rest,
therefore h2 + u2 - z2 has to shares on 2 without the rest:
h2 + u2 - z2 =2n2, where n2 - natural number.
Thus, is necessary for performance of the equation (1) performance
of system of the equations: x + at - z = 3n1
x2 + u2 - z2=2n2
x3 + u3 =z3
we Will consider Fermat`s equation of the fourth degree: h4 + u4 =z4 (4)
I we will transform it: x (h3 - x) + at (u3 - at) - z (z3- z) + h2 + u2 - z2 =0, where
x (h3 - x); at (u3 - at); z (z3-z) share on 3 without the rest, therefore h2 + u2 - too has to share
of z2 on 3 without the rest, i.e. h2 + u2 - z2 = 3k2 (5)
where k2 - natural number. we Will transform
(4) in the following look: h2 (h2 - x) + u2 (u2 - at) - z2 (z2 - z) + h3 + u3
- z3 =0, h2 (h2 - x); u2 (u2 - at); z2 (z2-z) share on 2 without the rest,
therefore and h3 + u3 - z3 has to share on 2 without the rest, i.e.
h3 + u3 - z3 =2k3, (6)
where k3 - natural number. Let`s transform (6) to a type
of x3 - x + u3 - at - (z3- z) + x + at - z = 2k3
h3 Here - x; u3 - at; (z3- z) share on 3 without the rest, therefore x + at - z - 2k3 shares on 3 without the rest, i.e. x + at - z - 2k3 =3k or x + at - z = k1 where the natural number of k1 = 2k3 +3k
Thus, is necessary for performance of the equation (4) performance of system of the equations: x + at - z = by k1
x2 + u2 - z2 =3k2
h3 + u3 - z3 = 2k 3
of x4 + u4 =z4
- 2 -
will Similarly be transformed the equations of other degrees. For the equation of fifth
of degree: h5 + u5 =z5 (7a)
transformation will be such: h5 - x + u5 - at - (z5- z) + x + at - z = 0 (8)
h3 (h2 - x) + u3 (u2 - at) - z3 (z2-z) + h4 + u4 - z4 = 0 (9)
h2 (h3 - x) + u2 (u3 - at) - z2 (z3-z) + h3 + u3 - z3 =0 (10)
From (8) we receive
: x + at - z =5m1, from (9) we receive h4 + u4 - z4 = 2m4, from (10) we receive h3 + u3 - z3 = 3 m3. From the last we receive the equation of the second
degree as follows: h3 - h2 + u3 - u2 - (z3 - z2) + h2 + u2 - = 3 m3 or
x (h2 - x) + at (u2 - at) - z (z2 - z) + h2 + u2 - z2 = 3 m3 (11)
From (11) we receive z2 that h2 + u2 - z2 - 3 m3 share without the rest on 2, i.e. x2 + u2 - z2 - 3 m3 =2m, or h2 + u2 - z2 = sq.m where natural number of sq.m = 2 m +3m3.
Thus in order that the equation was carried out (7a) is necessary that the system of the equations x + by y-z =5m1
h2 + u2 - z2 = in sq.m
h3 + u3 - z3 =3m3
h4 + u4 - z4 = 2m4
x5 + u5 =z5
System of the equations of a look was carried out x + at - z = v1
x2 + y2 - z2 = v2 (12)
h3 + u3 - z3 = v3
h4 + u4 - z4 = v4
can be received from Fermat`s equation of any degree. Always there will be such number and that r - and = would will be a prime number, then from
x and (hb - x) + ua (u6 - at) - za (z6-z) + by xa+1 + ya+1 - za+1 = 0
is received that x a+1 + at a+1 - z a+1 would be divided into number without the rest, i.e.
ha+1 + ua +1 - za+1 = bn (13)
where n - natural number. Similarly lowering degree of the equation we will receive system of the equations (12) for any equation of a look hp + Ur =zp.
hr + can replace Any equation with yp=zp with system of the equations both with higher degree, and with lower degree, and the quantity of the equations in system is not limited.
Proof No. 1 of the theorem of Fermat:
From h3 + u3 = z3 we receive the equation of higher degree, for example
(h3 - h5) + (u3 - u5) - (z3-z5) + + u5 - z5 = 0
of x2 (h3 - x) + u2 (u3 - at) - z2 (z3 - z) shares x5 on 3 without the rest, therefore
h5 + u5 - z5 shares on 3 without the rest, i.e.
h5 + u5 - z5 = 3n5. (13a)
From (13a) we receive the equation of the third degree h3 + u3 - z3 = B3 where B3 natural number not equal to zero. h3 + u3 = z3 is necessary for performance of the equation that there was a decision of system of the equations of a look
- 3 -
of x3 +y3 - z3 = B3 (13b) of
h3 + u3 = z3
that is impossible. Similarly for any equation of degree r xp +yp =zp
we receive the equation of a type of xp +yp - zp = BP and solving the system of the equations containing
of xp +yp - zp = BP
of xp +yp =zp
we draw a conclusion that there is no decision of this system, therefore the equation of xp +yp =zp has no decision and.
2. Proof No. 2 of the theorem of Fermat.
Lemma. + y3 = needs z3 for performance of the equation of the third degree of x3 that cubes of x3, y3, z3 were formed from identical cubes of V that their quantity was an integer and by quantity corresponded to numbers of n3, n9, n27 and so on (generally - (np)3 = n3p where n and r - natural numbers).
of Vh3R + Vh3D = Vh3F, Vh3R =n, Vh3D =m, Vh3F =k
Vn3 +Vm3 = Vk3, (14)
Where Vn3 = x3, Vm3 = y3, Vk3 = z3.
Let h3 - a cube of V1, y3 - a cube of V2, z3 - V3 cube. Let`s enclose V2 in V3 (we will subtract from V3 V2).
the Remained volume (V3 - V2) has to be equal to V1 volume. If the total amount
(crossing of V3 and V2) consists of n3p of quantity of small cubes of V, then the difference of volumes (V3 - V2) has to contain m3k of volumes of V in order that it was possible to construct V1 cube of these cubes, that is for performance of a condition
h3 + u3 = z3. Reducing (14) by V, we will receive n3 + m3 = k3 for which performance V1s3 needs V1 r3 + = to V1t3,
where to V1 r3 = to n3, V1s3 = to m3, V1t3 = to k3.
Reducing the equation (15) by V1 we will receive r3 + s3 = t3.
can So continue infinite number of times. From this it follows that if the equation h3 + u3 = z3 has at least one decision, then it will have infinitely many decisions. The equation of a look
of xp +yp =zp (15a)
can be presented in the form (xp/3) 3 + (yp/3) 3 = (zp/3) 3 (16)
- 4 -
which can be considered as cubic with the parties of a cube hr / 3; Ur / 3; zp/3. That the equation (15a) had the decision, it is necessary that the cube (hr / 3) 3 contained n3p of cubes of V, t. e. (hr / 3) 3 = V n3p; the cube (Ur / 3) 3 contained m3k of cubes of V, i.e.
(Ur / 3)3 = V m3k; the cube (zp/3) 3 contained r3f of cubes of V, i.e. (zp/3) 3 =Vr3f.
Then (16) it is transformed to a look
V n3p + V m3k = V r3f (16a) by
Reducing by V, we will receive
of n3p + m3k = r3f (16b) of
which after substitution of np = x; mk = at; rf = will be transformed by z
to Fermat`s equation of the third degree h3 + u3 =z3 which has no decisions, therefore, has no decisions and (15a). Theorem: The look equation hr + uk =zf for natural numbers x, at, z, r, to, f and at r, to, f> 2, is not carried out. in (16b) we will make by
For the proof substitution of n3 = x; m3 = at; r3 = z we will also receive
hr + uk = zf (16v) of
As the initial equation is not carried out (15a), also the equation is not carried out (16v).
the Proof No. 3
If is not carried out system of the equations (12), or at least two of its equations have no common decision, then the equation hr + Ur = will not be carried out by zp in natural numbers.
In system of the equations (12) we will consider the equation
of x2 + u2 - z2 = v2 or h2 + u2 =z2 + v2 = U. It has to have the decision in
natural numbers. Roots of this decision are known:
x =m2 - n2
at = 2mn
U = m2 +n2
the Equation h4 + u4 - z4 = v4 can be presented in the form (h2) 2 + (u2) 2 = (z2) 2+ v4
= (V2)2 which has to be carried out in natural numbers as the square
of x2=m2 - n2
u2 = 2mn
V2 = m2 + n2
Then h2 = x; u2 = at; V2 = U = (z4 + v4) 1/2 = (z2 + v2) ½ from which it is visible that x =1= at. At these values x, at xp equation + = zp cannot be carried out by yp in natural numbers.
Proof No. 4. Graphic.
- 5 -
From drawing it is visible that in the section of the plane x + at - z = v1 and an odnopolostny
hyperboloid h2 + u2 - z2 = v2 will be an ellipse. The surface h4 + u4 - z2 = v4 is symmetric to
concerning axis Z, as well as a surface h2 + u2 - z2 = v2. They can be crossed in the plane, perpendicular axis Z. In section there will be a circle. This circle can be crossed with an ellipse in one or two points. But it agrees to zp
proved in the item 2 the equation hr + Ur = cannot have the limited number of decisions, therefore the system of the equations (12) has no decisions and the equation hp + Ur = zp has no decision.
5. Proof No. 5.
we Will consider the equation
hkh + uu = zz (20)
easy to check substitution that it is not carried out for any natural numbers.
we Will present it in the form
(xkh/3) 3+ (uu / 3) 3 = (zz/3) 3
According to the item. 2 it will be transformed to a look
V n3p + V m3k = V r3f (21)
Reducing by V and doing replacement of n3 = to L; m3 = Q; k3 = W we will receive the equation
of Lp + Qr = Wf (22)
Which is not carried out as the initial equation is not carried out (20). Doing to
replacement in (21): np = L; mr = Q; kf = W we will receive
L3 + Q3 = W3
Which is not carried out as well as the initial equation (20).
Lp Equation + = Wp
is not carried out by Qp as is a special case of the equation (22)
22 follows From the equation that Fermat`s theorem is fair also at different values of exponents.