## When the product of any two of (a,b,c,d) is one less an integer squared

$a, \; b, \; c, \; d$   are positive integers such that

$1 \; + \; a \, b$
$1 \; + \; a \, c$
$1 \; + \; b \, c$
$1 \; + \; a \, d$
$1 \; + \; b \, d$
$1 \; + \; c \, d$

are all square numbers.

Here’s a simple parameterization :

$a \; = \; n \; - \; 1$
$b \; = \; n \; + \; 1$
$c \; = \; 4 \, n$
$d \; = \; 4 \, n \, (4n^2 - 1)$

$1 \; + \; a \; b \; = \; 1 + (n - 1) \;(n + 1) = n^2$

$1 \; + \; a \; c \; = \; 1 + 4 \; n \;(n - 1) = (2 \; n - 1)^2$

$1 \; + \; b \; c \; = \; 1 + 4 \; n \;(n + 1) = (2 \; n + 1)^2$

$1 \; + \; a \; d \; = \; 1 + (n - 1) \;(4 \; n \; (4 \; n^2 - 1)) = (4 \; n^2 - 2 \; n - 1)^2$

$1 \; + \; b \; d \; = \; 1 + (n + 1) \;(4 \; n \; (4 \; n^2 - 1)) = (4 \; n^2 + 2 \; n - 1)^2$

$1 \; + \; c \; d \; = \; 1 + (4 \; n) \;(4 \; n \; (4 \; n^2 - 1)) = (8 \; n^2 - 1)^2$

Here’s parameterization #2 ;

$a \; = \; m$
$b \; = \; n \, (m \, n +2)$
$c \; = \; (n + 1) \, (m \, n + m + 2)$
$d \; = \; 4 \, (m \, n + 1) \, (m \, n + m + 1) \, (m \, n^2 + m \, n + 2 \, n + 1)$

$1 + a \, b = 1 + m \, n \, (m \, n + 2) = (m \, n + 1)^2$

$1 + a \, c = 1 + m \, (n+1) \, (m \, n + m + 2) = (m \, n + m + 1)^2$

$1 + b \, c = 1 + n \, (m \, n +2) \,(n+1) \, (m \, n + m + 2) = (m \, n^2 + m \, n + 2 \, n + 1)^2$

$1 + a \, d$
$= 1 + m \, (4 (m \, n + 1) (m \, n + m + 1) \, (m \, n^2 + m \, n + 2 \, n + 1))$
$= (2 \, m^2 \, n^2 + 2 \, m^2 \, n + 4 \, m \, n + 2 \, m + 1)^2$

$1 + b \, d$
$= 1 + n \, (m \, n +2) \,(4 (m \, n + 1) \, (m \, n + m + 1) \, (m \, n^2 + m \, n + 2 \, n + 1)))$
$= (2 \, m^2 \, n^3 + 2 \, m^2 \, n^2 + 6 \, m \, n^2 + 4 \, m \, n + 4 \, n + 1)^2$

$1 + c \, d$
$= 1 + (n + 1) \, (m \, n + m + 2) \,(4 \, (m \, n + 1) \, (m \, n + m + 1) \, (m \, n^2 + m \, n + 2 \, n + 1))$
$= (2 \, m^2 \, n^3 + 4 m^2 \, n^2 + 2 \, m^2 \, n + 6 m \, n^2 + 8 \, m \, n + 2 \, m + 4 \, n + 3)^2$