Lemma.
1). If the rational point Ф* ( x*/z0, y*/z0) lies on the ray ƛ 0, then the length of the radius vector R* of this point is a rational number, (x*, y* are natural numbers).
2). If the rational point Ф* ( x*/z, y*/z) lies on the ray ƛ*, then the length of the radius vector R* is an irrational number (x*, y*, z* are natural numbers).
Let us prove the first part of the lemma.
Let P0 (x0/z0, y0/z0) be a point with rational coordinates that belongs to the unit circle (P0 ϵ S2 =1). Then the coordinates of this point satisfy the equation x20 + y20 = z20, (6) where x0, y0, z0 are natural numbers. Let the point Ф* ( x*/z0, y*/z0) with rational coordinates lie on the ray ƛ 0 passing through the point P0, (Ф* ϵ S2 =1).
Then ƛ 0 = y0/ x0 = y*/ x* and the length of the radius vector of point Ф* will be equal to R* = √((x*/z0)2 + (y*/z0)2) = x*/ z0 √(1+ ƛ 02) = x*/ z0 √(1+(y0/x0)2) = x*/ x0 – rational number.
2. The proof of this formula is given in Appendix 2.
Let us prove the second part of the lemma.
Let the point Ф*(x*/z*, y*/z*) with rational coordinates lie on the ray ƛ * which intersects the unit circle at the irrational point P* (u* ,v*), u*= m/k*, v* = n/k*.. Since ƛ * = y*/x* is a rational number, then the numerators of the fractions u* , v* are rational numbers: ƛ * = n/m. Whence it follows that the denominator k* = √(m2+n2) is an irrational number. The length of the radius vector of point Ф* will be equal to
R* = √( (x*/z*)2 + (y*/z*)2) = x*/z* √(1+(y*/x*)2) = x*/z* √(1+(n/m)2) = x*k*/z*m is an irrational number.
Thus, the length of the radius vector of a rational point on the ray ƛ0 (passing through the rational point of the unit circle) is a rational number, and the length of the radius vector of a rational point lying on the ray ƛ* (intersecting the unit circle at the irrational point) is the number irrational.Докажем вторую часть леммы.
Let x0, y0, z0 be a Pythagorean triple of numbers with pairwise coprime numbers. Such Pythagorean triplets of numbers are called primitive. Then on the grid in the unit square for the natural number z0 on the unit circle there is a Pythagorean point P0 (u0, v0), where u0 = x0/ z0, v0 = y0/ z0. The ray ƛ0 passing through the point P0 does not contain other grid nodes for z0. Let's prove it.
Let the nodal point Ф*(x*/z0, y*/z0) of the grid for z0 lie on the ray ƛ 0. Then y* = ƛ 0 x* = (y0/x0) x* (7)
Since x0 and y0 are relatively prime numbers, then y* will be an integer if x* is a multiple of x0, that is, x x* = m x0 where m is a natural number, m > 1. Then y* = my0 the coordinates of the point Ф* will be: u* = m x0 / z0, v* = m y0 / z0. It follows that the value of the radius vector of the point Ф* will be R* = m, where m is an integer. Since the largest value of the length of the radius vector of any point inside the unit square is less than √2, the point will be outside the unit square.
So, if the Pythagorean triple of numbers ( x0 , y0 , z0 ) is primitive, then on the Pythagorean ray ƛ0, except for the Pythagorean point P0 on the unit circle, there are no other nodal points of the grid. Then on the grid in the unit square for a natural number z0, all Fermat curves intersect the ray ƛ0 not at nodal points and, therefore, according to equation (1), there will be no integer solutions at these points. Then the transformed equation (3) will have irrational solutions for all n >2:
un = 1 /n√(1+ ƛ 0n ) = x0 / n√( x0n +y0n ), vn = ƛ 0 /n√(1+ ƛ 0n ) = y0 / n√( x0n +y0n )
Note that according to formula (8), the coordinates of the points of the Fermat curves on the ray ƛ 0 are formed from the coordinates of the Pythagorean point P0 (x0 / z0, y0 / z0) and, therefore, do not depend on other partitions of the unit square (obviously, the coordinates of the points do not depend at all on the partition of the unit square).
A comment.
On the one hand, the lemma states that on the Pythagorean ray ƛ 0 the length of the radius vector of the point Ф* with rational coordinates R* is a rational number. On the other hand, on a grid for z0 on which the point P0 on the unit circle is a nodal point (that is, the unit square for a given partition is integerly decomposed over the cells of the partition into the sum of two squares), there are no other nodal points on the ray ƛ 0 and we draw the conclusion (which has yet to be formally proven) that all Fermat curves intersect the Pythagorean ray ƛ 0 at irrational points. Let's give an example.
Example 1.
z0 = 5, P0 (3/5 , 4/5), ƛ 0 = 4/3. With such a partition (partition step h =1/5), the unit square can be decomposed into the sum of two squares with sides 3h and 4h. It is obvious that there is an infinite number of rational points on the ray, for example Ф* (1/2, 2/3). On a grid in a unit square for z = 6, this point will be a nodal point and lie on the ray ƛ0, Ф* (3/6, 4/6). The radius vector of this point is a rational number, R* =5/6. If we combine these two grids, that is, cover a unit square with a grid z1 = z0 , z = 30, then both points P0 (18/30, 24/30) and Ф* (20/30, 15/30) will be nodal points on this grid and are on the same ray ƛ0. The partitioning step of this grid is h = 1/30 and point P0 (18/30, 24/30) still splits the unit square into the sum of two squares with sides 18h and 24h.
In other words, let us make the assumption that some rational point Ф*n lies at the intersection of the Fermat curve Sn = 1 with the Pythagorean ray ƛ0, then we can always select a grid in the unit square in which both P0 and Ф*n will be nodal at the same time.
Let's return to the proof of the theorem. Since the numerators of the coordinates of the point Ф*n (un ,vn), (formulas (8)), are natural numbers, then solutions (8) will be irrational if we prove that n√ ( xn0+yn0) for n > 2 is an irrational number . Let's prove it.
Let for some natural number n, (n > 2 ), n√ ( xn0+yn0) = z10 /k – a rational number, where z10 and k are coprime natural numbers and the point Ф*n (kx0 / z10 , ky0 / z10 ) lies at the intersection of the Fermat curve Sn = 1 with the Pythagorean ray ƛ0. It is obvious that z10 /k < z0, since the length of the radius vector of the point Фn is greater than one, (Fig. 1)
Rn = √( (kx0 / z10 )2 + (ky0 / z10)2) = kz0 / z10 > 1 (9)
From the assumption that the point Ф n with rational coordinates lies on the Fermat curve Sn = 1, it follows that equation (1) must have an integer solution (kx0)n + (ky0)n = (z10)n (10)
From the last equation it follows that z10 must be divisible by k. But, by assumption, z10 and k are relatively prime numbers. Therefore, n√( (xn0 + yn0 ) cannot be a fractional rational number.
Let now n√(xn0+ yn0)be a natural number and z10 –z0. Then on the grid in the unit square for the natural number z10, the point Фn (xn0/ z10, y0/ z10) is a nodal point and lies on the Fermat curve Sn = 1, according to the proposal.
Obviously, for Pythagorean triples of numbers (x0, y0, z0), for which z0 = y0 +1 (for example, (3,4,5), (5,12,13), etc.), n√(xn0+ yn0) cannot be an integer, since in this case the inequalities must be satisfied.
y0 < n√(xn0+ yn0) < y0 +1 (11)
In other cases, when z0 > y0 +1, for example (20, 21, 29) from the assumption that the point Фn (x0/ z10, y0/ z10) lies on the Fermat curve Sn = 1, it follows that this point will be a nodal point on grid in the unit square for a natural number z10 that is not a multiple of z0 , (z0 > z10 and z0 is a prime number). According to formula (8), with n =2, the solution to equation (3) must be rational, and equation (1) must be integer. But on the grid in the unit square for z10, the point P0 will not be a nodal point and, therefore, with n = 2 at this point, equation (1) will not have an integer solution.
Therefore n√(xn0+ yn0) is an irrational number.
Proving that n√(xn0+ yn0) is irrational is not necessary. This follows from the previously proven statement that on the Pythagorean ray ƛ0 on the grid in the unit square for z0, except for the point P0 on the unit circle, there are no other nodal points. This is equivalent to the statement that for a right triangle with legs x0 and y0, the sum of legs to any other power n > 2 will not be a rational number (or rather an integer).
From the irrationality of n√(xn0+ yn0) for n > 2 it follows that the length of the radius vector of the point Фn lying at the intersection of the Fermat curve Sn = 1 with the Pythagorean ray ƛ0 is also an irrational number.
Rn = √(u2n+ vn2) = √( (x0/√(x0n+ y0n))2 + (y0/√(x0n+ y0n))2) = z0/n√(x0n+ y0n) (12)
It is obvious that there are infinitely many rational points on the ray ƛ0 and each of them, with an appropriate choice of the partitioning step of the unit square h + 1/z, can become a node on the grid for a natural number z. According to the lemma, the length of the radius vector of the point Ф* on the ray ƛ0 is a rational number. Consequently, for any partitions of the unit square, on the Pythagorean ray ƛ0 the inequality R* ≠ Rn holds, (13) where Rn is the length of the radius vector of the point Фn of the Fermat curve Sn = 1, and R* is the length of the radius vector of any rational point on the beam ƛ0.
