## √(a + b/c) = a √(b/c) — Part 2

Can you find positive integer triples   $(a, b, c)$,   with   $b$   and   $c$   cubefree and   $(b, c) = 1$

$\sqrt [3]{a + b/c} \; = \; a \: \sqrt [3]{b/c}$

If   $b = a$,   and   $c = a^3 - 1$

The condition   $b$   and   $c$   are cubefree   is not always satisfied

Here are the first few examples,

b = c   cubefree numbers:
2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 25, 26, 28, 29, 30

(2,2,27),   (3,3,26),   (4,4,63),   (5,5,124),   (7,7,342)
a = b = 9   —->   c = 728   is not a cubefree number
a = b = 10   —>   c = 999   is not a cubefree number
(11,11,1330),   (12,12,1727),   (13,13,2196),   (14,14,2743),   (15,15,3374),
a = b = 17   —>   c = 4912   is not a cubefree number
a = b = 18   —>   c = 5831   is not a cubefree number
a = b = 19   —>   c = 6858   is not a cubefree number
(20,20,7999),   (21,21,9260),   (22,22,10647), (23,23,12166)
a = b = 25   —>   c = 15624   is not a cubefree number
(26,26,17575)
a = b = 28   —>   c = 21951   is not a cubefree number
(29,29,24388),   (30,30,26999)

The necessary condition:

$a-1, \; a, \; a+1$   are all cubefree;   $b = a$,   $c = a^3 - 1$   is a cubefree number

is not always satisfied

E.g.

(1,2,3),   (2,3,4),   (3,4,5),   (4,5,6),   (5,6,7)

(9,10,11)   —>   c = 999   is not a cubefree number

(10,11,12),   (11,12,13),   (12,13,14),   (13,14,15),

(17,18,19)   —>   c = 5831   is not a cubefree number

……………………

math grad - Interest: Number theory
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### 3 Responses to √(a + b/c) = a √(b/c) — Part 2

1. Paul says:

In this case a = b and c = ((a^3 * b) – b)/a
Format {a, b, c}
giving:-
{2,2,7}
{3,3,26}
{5,5,124}
{6,6,215}
{7,7,342}
{10,10,999}
{11,11,1330}
{13,13,2196}
{14,14,2743}
{15,15,3374}
{17,17,4912}
{19,19,6858}

Paul.

• benvitalis says:

The condition
$b$   and   $c$   are cubefree numbers
is not always satisfied
E.g. {10,10,999}; 999   is not a cubefree number

2. benvitalis says:

$c \; = \; ((a^3 \: b) \: - \: b)/a$
$a \; = \; b$
then   $c \; = \; a^3 - 1$