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Konstantine Zelator

  1. The Rational Number n/p as a sum of two unit fractions.

    Пт, 03/02/2012 - 15:01 — SomNambul
    Authors: Konstantine Zelator
    Subjects: General Mathematics
    Abstract

    In a 2011 paper published in the journal "Asian Journal of Algebra"(see
    reference[1]), the authors consider, among other equations,the diophantine
    equations 2xy=n(x+y) and 3xy=n(x+y). For the first equation, with n being an
    odd positive integer, they give the solution x=(n+1)/2, y=n(n+1)/2. For the
    second equation they present the particular solution, x=(n+1)/3,y=n(n+1)/3,
    where is n is a positive integer congruent to 2modulo3.

  2. An ancient Egyptian problem:the diophantine equation 4/n=1/x+1/y+1/z, n>or=2.

    Втр, 12/15/2009 - 16:01 — SomNambul
    Authors: Konstantine Zelator
    Subjects: General Mathematics
    Abstract

    From the Rhind Papyrus and other extant sources, we know that the ancient
    Egyptians were very iterested in expressing a given fraction into a sum of unit
    fractions, that is fractions whose numerators are equal to 1. One of the
    problems that has come down to us in the last 60 years, is known as the Erdos-
    Strauss conjecture which states that for each positive integer n>1; the
    fraction 4/n can be decomposed into a sum of three distinct unit fractions.
    Since 1950, a numberof partial results have been achieved, see references [1]-
    [8]; and also [10] and[11].

  3. A cornucopia of pythagorean triangles.

    Пт, 10/02/2009 - 09:41 — SomNambul
    Authors: Konstantine Zelator
    Subjects: General Mathematics
    Abstract

    Consider two circles, externally tangential,and with integer radii R1, R2;
    and with R1>R2.The two circles have three tangent lines in common, one of them
    being T1T2. If M is the midpoint of T1T2, and K the point of intersection of
    the lines C1C2 and T1T2;then 16 right triangles are formed(C1 and C2 are the
    two circle centers), see Figure 1.In Section 6 of this paper, we find the
    precice form the two integers R1 and R2 must have, in order that the sixteen
    aforementioned right triangles be Pythagorean.

  4. Pythagorean Triangles with Repeated Digits-Repeated Bases.

    Чт, 08/27/2009 - 13:23 — SomNambul
    Authors: Habib Muzaffar, Konstantine Zelator
    Subjects: General Mathematics
    Abstract

    In 1998, in the winter issue of the journal Mathematics and Computer
    education (see [1]), Monte Zerger posed the following problem. He had noticed
    the Pythagorean triple (216,630,666);(216)^2+(630)^2=(666)^2. Note that 216=6^3
    and 666 is the hypotenuse length. The question was then, whether there existed
    a digit d and a positive integer k(other than the above); such that d^k is the
    leglength of a Pythagorean triangle whose hypotenuse length has exactly k
    digits, each being equal to d. In 1999, F.Luca and P.Bruckman, answered the
    above question in the negative.

  5. Pythagorean Triangles with Repeated Digits-Repeated Bases.

    Чт, 08/27/2009 - 13:23 — SomNambul
    Authors: Habib Muzaffar, Konstantine Zelator
    Subjects: General Mathematics
    Abstract

    In 1998, in the winter issue of the journal Mathematics and Computer
    education (see [1]), Monte Zerger posed the following problem. He had noticed
    the Pythagorean triple (216,630,666);(216)^2+(630)^2=(666)^2. Note that 216=6^3
    and 666 is the hypotenuse length. The question was then, whether there existed
    a digit d and a positive integer k(other than the above); such that d^k is the
    leglength of a Pythagorean triangle whose hypotenuse length has exactly k
    digits, each being equal to d. In 1999, F.Luca and P.Bruckman, answered the
    above question in the negative.

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