Find a,b,c so that a/b + b/c + c/a is an integer

S = a/b + b/c + c/a an integer, a ≠ 0, b ≠ 0 & c ≠ 0

(a^2 c + a b^2 + b c^2)/(a*b*c)

Trivial solution: a = b = c ⇒ S = 1 + 1 + 1 = 3

a = 2, b = 4 & c = 8 ⇒ S = 2/4 + 4/8 + 8/4 = 3

Even numbers of the form 2n, 4n & 8n ⇒ S = 2n/4n + 4n/8n + 8n/2n = 3

a = 2, b = 12, c = 9 ⇒ S = 2/12 + 12/9 + 9/2 = 6

a = 3, b = 126, c = 196 ⇒ S = 3/126 + 126/196 + 196/3 = 66

Solution found in 1931: a = 1539, b = 1369634, c = 129549924

1539/1369634 + 1369634/129549924 + 129549924/1539 = 84178

[To Be Continued]

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About benvitalis

math grad - Interest: Number theory
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2 Responses to Find a,b,c so that a/b + b/c + c/a is an integer

  1. Number of the form n,2n,4n

    S = n/2n + 2n/4n + 4n/n = 5

  2. benvitalis says:

    Right! Numbers of the form n, 2n, 4n –> n/2n + 2n/4n + 4n/n = 5
    n/2n + 2n/4n + 4n/n is 1/2 + 1/2 + 4
    We can find many other examples, where the sum of the first two fractions produce an integer if the third fraction is k an integer. For example,
    1/3 + 2/3 + k
    that is, the first fraction is m/n = 1/3, the second fraction n/p = 2/3

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