Monthly Archives: December 2015

(a,b); gcd(a,b) = 1; (a^2 – 5)/b and (b^2 – 5)/a are positive integers

  Does there exist an infinite numbers of pairs (a, b) satisfying the given conditions?                                  

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Two Primitive Pythagorean triples

    and     are two primitive Pythagorean triples The primitive Pythagorean triples     are characterized by    and    for some positive integers     and   ,   and that then   For the given Pythagorean … Continue reading

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Sum of a geometric series : 1 + r + r^2 + … + r^n = A^2

  The general form of a geometric sequence is   is the first term, and   is the factor between the terms (called the “common ratio”) Find a geometric series of 3 or more positive integers, starting with 1, such … Continue reading

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x^4 + y^4 = z^4 – N with (x,y,z) in arithmetic sequence

    with     and in arithmetic sequence.   There are solutions of the form:                                                       ……….    (1,   2,   3),   (7,   8,   9)   ……….    (2,   4, … Continue reading

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Average of 3^2, 4^2, 5^2, …, n^2 is itself a perfect square

    Average of 1^2, 2^2, 3^2, …, n^2 is itself a perfect square                                        

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x^4 + y^4 = z^2 ± 1

      Here are the solutions for all       Find solutions for,                                    

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Triangles (a,b,c),(m,n,p) and (√(a^2+m^2),√(b^2+n^2),√(c^2+p^2))

    Prove that if     and     are the lengths of the sides of two triangles then ,        ,        are also the lengths of the sides of some triangle.             … Continue reading

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Average of 1^2, 2^2, 3^2, …, n^2 is itself a perfect square

  First few integers     such that the average of         is itself a perfect square                            

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Fibonacci num3ers : a surprising occurrence

  To find positive integers     such that                     are all squares   that is,                     Note that   It happens that the first few Fibonacci numbers can be used to … Continue reading

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Equation : A^2 + B^2 + 1 = C^2 + D^2 —- (Part 4)

                                            In   A^2 + B^2 + 1 = C^2 + D^2 —- (Part 1) and In   Equation : A^2 + B^2 + 1 = C^2 + D^2 —- (Part 2) We can write                         … Continue reading

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