## (ab)(a + b)(a^2 + ab + b^2) = (cd)(c + d)(c^2 + cd + d^2)

$a \, b \, (a + b) \,(a^2 + a \, b + b^2) \; = \; c \, d \, (c + d) \,(c^2 + c \, d + d^2)$

where   $a, b, c, d$   are integers

Here’s one solution:

$(a, \: b, \: c, \: d) \; = \; (105, \: 1153, \: 455, \: 582)$

$(105\times 1153)(105 \; + \; 1153)(105^2 \; + \; 105\times 1153 \; + \; 1153^2)$
$= \; (455\times 582)(455 \; + \; 582)(455^2 \; + \; 455\times 582 \; + \; 582^2)$
$= \; 222585961555230$

Note that

$x^5 \; + \; y^5 \; - \; (x+y)^5 \; = \; a^5 \; + \; b^5 \; - \; (a+b)^5$
$-5 \, x \, y \, (x+y) \, (x^2 + x \,y + y^2) \; = \; -5 \, a \, b \, (a+b) \, (a^2+ a \, b + b^2)$
$x \, y \, (x+y) \, (x^2 + x \, y + y^2) \; = \; a \, b \, (a+b) \, (a^2 + a \, b + b^2)$

$(a, \: b, \: c, \: d) \; = \; (105, \: 1153, \: 455, \: 582)$

$105^5 \; + \; 1153^5 \; - \; (105+1153)^5 \; = \; 455^5 \; + \; 582^5 \; - \; (455+582)^5$

$(105\times 1153)(105 \; + \; 1153)(105^2 \; + \; 105\times 1153 \; + \; 1153^2)$
$= \; (455\times 582)(455 \; + \; 582)(455^2 \; + \; 455\times 582 \; + \; 582^2)$
$= \; 222585961555230$

math grad - Interest: Number theory
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### 2 Responses to (ab)(a + b)(a^2 + ab + b^2) = (cd)(c + d)(c^2 + cd + d^2)

1. Paul says:

Here are a few more, Format {{a, b, etc},{c, d, etc}}

{{210, 2306, 7122750769767360}, {910, 1164, 7122750769767360}}
{{315, 3459, 54088388657920890}, {1365, 1746, 54088388657920890}}
{{420, 4612,227928024632555520} ,{1820, 2328, 227928024632555520}}

Paul.

• benvitalis says:

Giving us an equal sum of three 5-th powers