arXiv subjects

Konstantine "Hermes" Zelator

Publications and source records attributed to Konstantine "Hermes" Zelator.

5 recordsLinked to original sources

arXiv2008

On a Theorem on sums of the form 1+2^(2^n)+2^(2^n+1)+...+2^(2^n+m) and a result linking Fermat with Mersenne numbers

In his book "250 Problems in Elementary Number Theory", W.Sierpinski shows that the numbers 1+2^(2^n)+2^(2^n+1) are divisible by 21; for n=1,2,.... In this paper, we prove a similar but more general result.Consider the natural numbers of the form I(n.m)= 1+2^(2^n)+2^(2^n+1)+...+2^(2^n+m).In Theorem 1 we prove that for every odd integer N greater than 1, there exist infinitely many natural numbers n and m such that the integers I(n.m) are divisible by N. We give an explicit construction of the numbers n and m, for a given N. As an example, when N=31, and with n=4k and m=94+124i, the numbers I(n,m) are divisible by 31. A similar example is offered for N=(31)(7)=217. In Theorem 2, we prove a result pertaining to Mersenne numbers.There are also three Corollaries in this work, one of which deals with Fermat numbers.

math.GM↗

arXiv2008

Two Theorems on the structure of Pythagorean triples and some diophantine consequences

Even though four theorems are actually proved in this paper, two are the main ones,Teorems 1 and 3. In Theorem 1 we show that if a and be are odd squarefree positive integers satisfying certain quadratic residue conditions; then there exists no primitive Pythagorean triangle one of whose leglengths is equal to a times an integer square, while the other leglength is equal to b times a perfect square. The family of all such pairs (a,b) is slightly complicated in its description. A subfamily of the said family consists of pairs (a,b), with a being congruent to 1, while b being congruent to 5 modulo8; and also with both a and b being primes, and with a being a quadratic nonresidue ofb(and so by the quadratic reciprocity law, b also being a nonresidue of a). Theorem 3 is similar in nature, but less complicated in its hypothesis. It states that if p and q are primes, both congruent to 1 modulo4, and one of them being a quadratic nonresidue of the other.Then the diophantine equation, p^2x^4 + q^2y^4 = z^2, Has no solutions in positive integers x, y, and z, satisfying (px, qy)=1.

math.GM↗

arXiv2008

A little noticed right triangle

Given a right triangle ABC, with the ninety degree angle at A; consider the triangle O1OO2.Where the point O is the midpoint of the hypotenuseBC(and so the center of the triangle ABC's circumcircle), the point O1 being the triangle AOB's circumcenter, and the point O2 being the triangle AOC circumcenter. The triangle O1OO2 is actually a right triangle similar to ABC. In the early part of this work, some geometrical formulas are derived, including some lengths of segments from a nearby trapezoid. After that, the focus changes. We examine the case when ABC is a Pythagorean triangle, and find the precise conditions under which the triangle O1OO2 is also Pythagorean. A numerical table of values is also offered at the end of the paper.

math.GM↗

arXiv2008

A solution to a problem and the Diophantine equation X^2+bX+c=Y^2

We prove that for given integers b and c, the diophantine equation x^2+bx+c=y^2, has finitely many integer solutions(i.e. pairs in ZxZ),in fact an even number of such solutions(including the zero or no solutions case).We also offer an explicit description of the solution set. Such a description depends on the form of the integer b^2-4c. Some Corollaries do follow. Furthermore, we show that the said equation has exactly two integer solutions, precisely when b^2-4c= 1,4,16,-4,or-16. In each case we list the two solutions in terms of the coefficients b and c.

math.GM↗

arXiv2008

Certain Properties of Pythagorean Triangles involving the interior diameter, and the exterior diameters

There are four characteristic circles for each triangle on a plane. All for are tangential to the three straight lines containing the triangles' three sides. Three are exterior circles, the fourth is the in-circle. When the triangle is Pythagorean, the four diameters are integers. Consider a Pythagorean triangle with the property that one leglength is a perfect(or integer)square, and with one of the four diameters also a integer square.Of the eight resulting combinations, we prove that only six are possible or can occur. We then completely parametrically describe the six families; each corresponding to one of the six combinations.

math.GM↗