## √(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;   $b = a$,     $c = a^2 - 1$

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

### 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.

• benvitalis says:

I get different results

• Paul says:

yes some are the same, seems you get some I don’t and I get some you don’t.