√(a + b/c) = a √(b/c) — Part 1

 
 
Find all positive integer triples   (a, b, c),   with   b   and   c   squarefree and   (b, c) = 1

\sqrt {a + b/c} \; = \; a \: \sqrt {b/c}

 
Then   a \, c \; = \; b \,(a^2 - 1).
Since   (a, \: a^2 - 1) \; = \; (b, \: c) \; =  \;1,
we must have   a|b   and   b|a,   so   a = b   and   c = a^2 - 1.
Hence   a,   like   b,   is squarefree, and so are   a+1   and   a-1  
since their product   c   is squarefree

Thus we have the necessary conditions:

a-1, \; a, \; a+1   are all squarefree;   b = a,     c = a^2 - 1

 
 
sqrt-equation-1

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
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3 Responses to √(a + b/c) = a √(b/c) — Part 1

  1. Paul says:

    There are an infinite number of triples where :-

    a = b = 2^(n+1) – 2 and
    c = 4^(n+1) – 2^(n+3) + 3

    For all n, here are the first 10, format {a, b, c}.

    {2,2,3}
    {6,6,35}
    {14,14,195}
    {30,30,899}
    {62,62,3843}
    {126,126,15875}
    {254,254,64515}
    {510,510,260099}
    {1022,1022,1044483}
    {2046,2046,4186115}

    Paul.

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