arXiv subjects

Konstantine Zelator

Publications and source records attributed to Konstantine Zelator.

At least 19 recordsLinked to original sources

arXiv2013

The sum of the squares of p positive integers which are consecutive terms of an arithmetic progression: Always a non-perfect square when p(a prime)=3 or p is congruent to 5 or 7 modulo 12

In a paper published by this author in www.academia.edu(see reference[3]), it was established that there exist no three positive integers which are consecutive terms of an arithmetic progression; and whose sum of squares is a perfect or integer square. In that paper, we made use of the 3-parameter formulas which describe the entire set of positive integer solutions of the 4-variable equation, $x^2+y^2+z^2= t^2$ (See reference [1]) In this work, we offer an alternative proof to the above result; a proof that uses only powers of 3 divisibility arguments. This is done in Theorem1, Section2. After that, in Proposition3(Section4) we use the Quadratic Reciprocity Law for odd primes, to establish that if p is a prime congruent to 5 or 7 modulo12; then 3 is quadratic non-residue of p. This then, plays a key role in proving Theorem 2, which postulates that if p is an odd prime congruent to 5 or 2 mod12; then, the sum of the squares any p natural numbers which are consecutive terms of an arithmetic progression; is a non-perfect square. Theorem3 is an immediate corollary of Theorem2 : for primes p congruent to 5 or 7 mod12; the sum of the squares of p natural numbers, consecutive terms of an arithmetic progression; cannot be a perfect square.

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arXiv2013

Integer Solutions, Rational solutions of the equations x^4+y^4+z^4 -2(x^2)(y^2)-2(y^2)(z^2)-2(z^2)(x^2)=n and x^2+y^4+z^4-2x(y^2)-2x(z^2)-2(y^2)(z^2)=n; And Crux Mathematicorum Contest problem CC24

The subject matter of this work are the two equations: x^4+y^4+z^4-2(x^2)(y^2)-2(y^2)(z^2)-2(z^2)(x^2)= n (1) And x^2+y^4+z^4-2x(y^2)-2x(z^2)-2(y^4)(z^4)= n (2) where n is a natural number. Contest Corner problem CC24, published in the May2012 issue of the journal Crux Mathematicorum(see reference[1]); provided the motivation behind this work. In Th.1, we show that eq.(1) if n=8N, N odd; then eq.(1) has no integer solutions; which generalizes problem CC24(the case n=24). We use Th.2, to find some rational solutions of eq.(1); which answers the second question in CC24. In Th.4, we show that if n= p, 4, or pq; where p and q are distinct primes. Then eq.(1)has no integer solutions. In Th.6, we determine all the integer solutions to (1), when n=p^2, p an odd prime. Theorems 7 through13, deal with equation (2). In Th.11, we determine all the integer solutions of eq.(2). Th.12 states that (2) has no integer solutions in n is congruent to 2 or 3 modulo4. Finally, in Th.13 we determine all the rational solutions of eq.(2).

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arXiv2013

Integer-Sided Triangles with integral medians

In this work, we prove that any triangle whose three sidelengths are integers, cannot have all of its three medians also having integral lengths.This is done in Proposition 2.In Section 5, we give precise(i.e.necessary and sufficient)conditions for a nonisosceles, integer-sided triangle to have two integral medians.In Section 3, we offer parametric descriptions of three special families of integer-sided triangles.The first family consists of all Pythagorean triangles whose medians to their hypotenuses is integral. The other two medians (to the two legs)of any Pythagorean triangle are irrational, a proof of this fact can be found in reference 4 of this paper.The second family consists of all integer-sided isosceles triangles, whose only integral median is the one contained between the two sides of equal length. Finally, the third family consists of all isosceles integer-sided triangles with the two medians of equal length;having integer length.

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arXiv2013

The Diophantine equation xy=z^n; for n=2,3,4,5,6; the Diophantine equation xyz=w^2; and the Diophantine system: xy=v^2 and yz=w^2

In this work, we accomplish three goals. First, we determine the entire family of positive integer solutions to the three- variable Diophantine equation, xy=z^2; for n=2,3,4,5,6. For n=2, we obtain a 3-parameter family of solutions; for n=3, a 5-parameter of solutions; likewise for n=4. For n=5, a 7-parameter family of solutions; and likewise for n=6. See Theorems 2 through 6 respectively. The second goal of this paper, is determining all the positive integer solutions of xyz=w^2. This is done in Theorem7; the solution set is described in terms of six independent parameters. Finally, in Theorem 8, we achieve our third goal: determining all the positive integer solutions of the 5-variable Diophantine system: xy=v^2 and yz=w^2. The solution set is expressed in terms of eight parameters. This paper contains a total of four references.

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arXiv2013

Integral points on rational curves of the form y=(x^2+bx+c)/(x+a); a,b,c integers

The subject matter of this work is the set of integral points(i.e. points with both coordinates integers) on the graphs of rational functions of the form f(x)=(x^2+bx+c)/(x+a), with a,b,c,being integers.Following the introduction, we establish Proposition1 in Section2. This proposition plays a key role in the proof of Theorem1 in Section5. Proposition1 is proved with the aid of Euclid's lemma and another well known result in number theory; see reference [1].In Sections3 and4, we focus on the special case b^2-4c=0; which implies b=2d and c=d^2, for some integer d. If d and a are distinct; then there are finitely many integral points, parametrically described in Results1 and2. Theorem1 in Sec.5 deals with the general case.Accordingly, if a^2-ab+c is not zero; there are exactly 4N distinct integral points parametrically described.Except in the cases where a^2-ab+c is a perfect square, or minus a perfect square; in which cases thare are exactly 4N-2 distinct integral points. Here N stands for the number of positive divisors, not exceeding the square root of the absolute value of a^2-ab+c. (divisors of that absolute value).The paper concludes with Th.2, which is a direct application of Th.1 in the cases where the above absolute value is 1, p, or p^2; p a prime.

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arXiv2012

Integral regular truncated pyramids with rectangular bases and the diophantine equation x^2+y^2+z^2= t^2

A regular truncated pyramid with rectangular bases;consists of two rectangular bases whose centers are orthogonally aligned with respect to the parallel planes containing their bases; and two pairs of congruent isosceles trapezoids(the four lateral faces). Thrre are six lengths involved:the larger base dimensions a and b; a>(or=)b. The smaller base dimensions c and d; c>(or=d). The height H, and the common length t of the four lateral faces. When a,b,c,d,H,t, and the volume V are all positive integers; we have an integral regular truncated pyramid with rectangular bases(see Definition 1 in the introduction). The two key geometric conditions that the above six lengths must satisfy are, a/b=c/d(see Section 3) and the equation, 4t^2= 4H^2+(a-c)^2+(b-d)^2 (*), derived in Section4. When H,a,c,b,d,t; are all positive integers. A modulo4 congruence shows that both the positive integers a-c and b-d; must be even. Consequently, equation (*) reduces to the equation, t^2= H^2+x^2+y^2 (**). All the positive integer solutions to (**) can be found, parametrically described, in reference [1]. Using the general positive integer solution to (**), we Proposition 1(Section7); which gives precise conditions that describe the set of all integral regular truncated pyramids with rectangular bases. In Proposition 2, we describe a 3-parameter family of such pyramids. In Proposition 3, we describe the square case: a=b> c=d.

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arXiv2012

Integral Triangles with one angle twice another, and with the bisector(of the double angle) also of integral length

Let ABC be a triangle with a,b,and c being its three sidelengths. In a 1976 article by Wynne William Wilson in the Mathematical Gazette(see reference[2]), the author showed that angleB is twice angleA, if and only if b^2=a(a+c). We offer our own proof of this result in Proposition1.Using Proposition1 and Lemma2, we establish Proposition 2: Let a,b,c be positive reals. Then a triangle ABC having a,b,c as its sidelengths can be formed if,and onlyif, b^2=a(a+c) and either c<(or equal to)a; or alternatively a<c<3a. Now, consider the case of integral triangles, that is; a,b, and c bieng positive integers.In 2002, in a paper published in the Mathematical Gazette(see[2]), author M.N.Deshpande provided two-parameter formulas that describe some integral triangles with (angle)B=2(angle)A. In Result2 in Section5, we offer 3-parameter formulas that describe the entire family of integral triangles ABC with angleA=2angleB. Using Result1, we then parametrically describe the entire family of integral triangles with angle A=2angleB; and with the bisector of angleB also of integral length. This is done in Reult2 in Section6. In Section7, we conclude this article with two closing remarks.

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arXiv2012

The Diophantine Equation arctan(1/x)+arctan(m/y)= arctan(1/k)

In the fall 2011 issue of the Journal'Mathematics and Computer Education', author Unal Hasan, in the one page article "Proof without Words", gives a purely geometric proof of the equality, arctan(1/3)+ arctan(1/7) = arctan(1/2) (1) (See reference [1]) Now consider the two-variable diophantine equation(x and y being positive integer variables), arctan(1/x) + arctan(m/y) = arctan(1/k) (2), where m and k are given or fixed positive integers with gcd(m,k^2+1)=1;and also with gcd(m,y)=1. Equality (1) then says that the pair (3,7)is a positive integer solution to (2) in the case m=1=k. We prove, in Theorem1(a,) that equation (2) has exactly N(number of positive divisors of k^2+1) distinct positive integer solutions (x,y), given by x=k+m(k^2+1)/d, y=km+d; d a positive divisor of k^2+1. As a result of Th.1, we list nine arctangent equalities in Section5 of this article, including inequality (1) above.

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arXiv2012

An Application of Ptolemy's Theorem:Integral triangles with a 120 degree angle and the bisector(of the 120degree angle)also of integral length

In one of the three 2010/2011 issues of the journal 'MathematicalSpectrum', this author gave a three-parameter description of the entire set of integral triangles(i.e. triangles with integer side lengths)and with a 120 degree angle.This entire set expressed as a union of four families, see reference[5]. In this work we describe, in terms of three parameters again, the set of all integral with a 120 degree angle, and whose bisectors of their 120 degree angles; is also of integral length. To do so, we use the well known historic theorem of Ptolemy for cyclic quadrilaterals, in conjunction with the general positive integer solution of the equation, 1/z=1/x +1/y; and of course in combination with the parametric description of the set of integral triangles with a 120 degree angle mentioned above,The final results of this paper are found in section8.

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arXiv2012

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

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. If in the above equations we assume n to be prime, then these two equations become special cases of the diophantine equation, nxy=p(x+y) (1), with p being a prime and n a positive integer greater than or equal to 2. This 2-variable symmetric diophantine equation is the subject matter of this article; with the added condition that the intager n is not divisible by the prime p. Observe that this equation can be written in fraction form: n/p= 1/x + 1/y(See [2] for more details) In this work we prove the following result, Theorem1(stated on page2 of this paper):Let p be a prime, n a positive integer at least2, and not divisible by p. Then, 1)If n=2 and p is an odd prime, equation (1) has exactly three distinct positive integer solutions:x=p, y=p ; x=p(p+1)/2, y=(p+1)/2 ; x=(p+1)/2, y=p(p+1)/2 2)If n is greater than or equal to 3, and n is a divisor of p+1. Then equation (1) has exactly two distinct solutions: x=p(p+1)/n, y=(p+1)/n ; x=(p+1)/n, y=p(p+1)/n 3) if n is not a divisor of p+1. Then equation (1) has no positive integer solution. The proof of this result is elementary, and only uses Euclid's Lemma from number theory,and basic divisor arguments(such that if a prime divides a product of two integers; it must divide at least one of them).

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arXiv2011

Five Exponential Diophantine Equations and Mayhem Problem M429

Crux Mathematicorum with Mathematical Mayhem, is a problem solving journal published by the Canadian Mathematical Society. In the March 2010 issue(see reference[1]) ,the following problem was proposed:Determine all positive integers a,b, and c such that a^(b^c)=(a^b)^c; or equivalently, a^(b^c)=a^(b^c). A solution by this author was published in the December2010 issue of Crux(see reference[2]). Accordingly, all such positive integer triples are the following:The triples of the form (1,b,c); with b, c any positive integers; the triples (a,b,1); a, b positive integers, with a being at least 2; and the triples of the form (a,2,2); a being a positive integer not equal to 1.These are then the positive integer solutions to the 3-variable exponential diophantine equation, x^(y^z)=x^(yz) (1) Motivated by mayhem problem M429, in this work we investigate for more 3-variable exponential diophantine equations: x^(y^z)=x^(z^y) (2), x^(y^z)=y^(xz) (3) x^(yz)=y^(xz) (4), x^(y^z)=z^(xy) (5) We completely determine the positive integer solution sets of equations (2), (3), and (4). This is done in Theorems2,3, and4 respectively. We also find three different families of solutions to equation (5); listed in Theorem5.

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arXiv2011

Integer roots of quadratic and cubic polynomials with integer coefficients

The subject matter of this work is quadratic and cubic polynomial functions with integer coefficients;and all of whose roots are integers. The material of this work is directed primarily at educators,students,and teachers of mathematics,grades K12 to K20.The results of this work are expressed in Theorems3,4,and5. Of these theorems, Theorem3, is the one that most likely, the general reader of this article will have some familiarity with.In Theorem3, precise coefficient conditions are given;in order that a quadratic trinomial(with integer) have two integer roots or zeros.On the other hand, Theorems4 and5 are largely unfamiliar territory. In Theorem4, precise coefficient conditions are stated; for a monic cubic polynomial to have a double(i.e.of multiplicity 2) integer root, and a single integer root(i.e.of multiplicity 1).The entire family of such cubics can be described in terms of four groups or subfamilies; each such group being a two-integer parameter subfamily. In Theorem5, a one-integer parameter family of quadratic trinomials(with integer coefficients) with two integer roots; is described.The parameter can take any odd positive integer values.

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arXiv2011

Reciprocal Properties of Pythagorean triangles

Let (a,b,c)be a Pythagorean triple with c being the hypotenuse length, and h being the altitude to the hypotenuse. Also, let v,k,l be positive integers with k and l being relatively prime.We say(Definition1 in this work)that the Pythagorean triple (a,b,c) has the reciprocal property R(v,k,l)if the positive integers a,b,c,v,k,and l satisfy the condition or equation, 1/a+1/b+v/h = k/l. The motivating force behind this work, is a problem that appeared in the journal, Crux Mathematicorum with Mathematical Mayhem. The said problem is Mayhem problem M390, and it appeared in the April2009 issue of the journal(see reference {1}). A solution to the same problem was published in the February 2010 issue(see {2}).Using the above definition, the probem can be stated as follows:Find all the Pythagorean triples that have the reciprocal property R(1,1,1). It turns out that the only such triple is (3,4,5).Our solution to this problem is found in Theorem2, part(iii).The results of this paper are expressed in Theorems2,3,4 and 5. Theorem2 contains five parts.Part(ii) states that there exists no Pythagorean triple which has the reciprocal property R(2,k,1). Part(iv) says that if k is at least2; then there exists no Pythagorean triple that has the reciprocal property R(1,k,l). Theorem3 states that if the positive integer v is such that the integers v-1 and v+1 are twin primes; then there is no Pythagorean triple that has the reciprocal property R(v,k,1).

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arXiv2011

Properties of proper rational numbers

This short article is aimed at educators and teachers of mathematics.Its goal is simple and direct:to explore some of the basic/elementary properties of proper rational numbers.A proper rational number is a rational which is not an integer. A proper rational r can be written in standard form: r=c/b,where c and b are relatively prime integers; and with b greater than or equal to 2. There are seven theorems, one proposition, and one lemma; Lemma1, in this paper. Lemma1 is a very well known result, commonly known as Euclid's lemma.It is used repeatedly throughout this paper, and its proof can be found in reference[1]. Theorem4(i) gives precise conditions for the sum of two proper rationals to be an integer.Theorem5(a) gives exact conditions for the product to be an integer. Theorem7 states that there exist no two proper rationals both of whose sum and product are integers.This follows from Theorem6 which states that if two rational numbers have a sum being an integer; and a product being an integer;then these two rationals must both be in fact integers.

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arXiv2010

Pythagorean triangles within Pythagorean triangles

In this work, we investigate the following question. Given a Pythagorean triangle BCA, with the right angle at C, let P be a point on the hupotenuse BA; and let D and E be the perpendicular projections of the point P onto the sides BC and CA respectively.When is either of the right triangles BDP and PEA Pythagorean? As it turns out, according to Theorem1, they are either both Pythagorean, or neither of them is.When they are both Pythagorean,a complete parametric description of these two triangles is given; in terms of the parameters that describe the Pythagorean triangle BCA. Later in the paper, we offer a complete analysis of three special cases:the case wherein the point P is the midpoint M of the hypotenuse BA; the case when P is the foot I of the 90 degree angle bisector;and the case in ehich the point P is the foot F of the perpendicular drawn from the vertex C to the hypotenuse. After that, some other cases are investigated as well.Specifically, we consider the case in which not only the triangles BDP and PEA are Pythagorean;but the four congruent right triangles DCP,DEP,DCE,and CEP,are Pythagorean as well.Precise conditions in order for this to occur, are given in Theorem7. Also,in Theorem 2, part(i), we show that if the triangle BCA is a primitive Pythagorean triangle, there exists no point P along the hypotenuse BA, for which both right triangles BDP and PEA are Pythagorean.

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arXiv2010

The Diophantine Equation x^n+y^m=c(x^k)(y^l), n,m,k,l,c natural numbers

The subject matter of this work is the diophantine equation x^n+y^m=c(x^k)(y^l), where n,m,k,l,c are natural numbers.We investigate this equation from the point of view of positive integer solutions.A preliminary examination of sources such as reference[1](L.E.Dickson's History of the Theory of Numbers, Vol.II) and [2](W.Sierpinski's Elementary Theory of Numbers) shows that little or no material can be found regarding this diophantine equation.Note that when c=1, (x,y)=(1,1) is a solution regardless of the values of the exponents n,m,k,and l. In Section3, five results from number theory are listed.The first four are well known and are stated without proof.Result5 is of central importance and it is used in the proofs of most of the nine theorems of this paper.We offer a detailed proof of Result5. The entire paper is organized according to eight cases. Here is a sample of two of the nine theorems. In Theorem2, we prove that if n<or=k and n<m<k+l, then the above has no positive integer solutions if c is not equal to 2;if c=2, then the above equation has the unique solution(x,y)=(1,1). In Theorem5, part(vi), we prove that if c=3, n-k=1,l=1,and m-n=1.Then the above equation has exactly two solutions:the (x,y)=(2,2),(2^k+1,2^k).

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arXiv2010

Pythagorean Boxes with Primitive Faces

In their paper "Pythagorean Boxes", Raymond A.Beauregard and E.R.Suryanarayan define the concept or notion of Pythagorean Rectangle as one with sidelengths and integer diagonal lengths(see [1]);they also introduce the concept of a Pythagorean Box as a rectangular three-dimensional parallelepiped whose edges and diagonals have integer lengths.As in that paper, the abbreviation PB will simply stand for "Pythagorean Box";also in this article,the abbreviated notion PR will stand for "Pythagorean Rectangle". In the Beauregard and Suryanarayan paper,it was shown that there exist infinitely many PB's with a square base and height equal to 1.In this paper,we present a method and formulas that generate infinitely many PB's that contain a pair of opposite(and hence congruent)PR's which are primitive;a PR is primitive if the four congruent Pythagorean triangles contained there in are primitive.There are three results in this paper.In Result1,we derive certain explicit conditions that a PB must satisfy, if it possesses two pairs of(opposite)primitive PR's.In Result2,we show that if similar conditions are satisfied then infinitely many PB's can be generated containing a pair of(opposite)PR's.In Result3, we prove that there exist no PB's with a square base and a face(and hence two faces)which is a primitive PR.

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arXiv2010

Implementing the Law of Sines to solve SAS triangles

By "solving a triangle", one refers to determining the three sidelengths and the three angles, based on given information.Depending on the specific information, one or more triangles may satisfy the requirements of the given information.In the SAS case, two of sidelengths are given, as well as the angle contained by the two sides.According to Euclidean Geometry, such a triangle must be unique. In reference [1], and pretty much in standard trigonometry and precalculus texts,the Law of Cosines is employed in solving a SAS triangle. In this work we use an alternative approach by using the Law of Cosines.In Section 2, we list some basic trigonometric identities and in Section 3 we prove a lemma which is used in Section4. In Section4, we demonstrate the use of the Law of Sines in solving an SAS triangle. In Section 5 we offer three examples in detail; the last one being more general in nature.

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