## (a,b,c,d) the product of any two positive integers is one less than a square

Given any four positive integers   $a, \; b, \; c, \; d$   such that the product of any two is one less than a square number

If we let

$a \; = \; n$
$b \; = \; n \; + \; 2$
$c \; = \; 4 \, n \; + \; 4$
$d \; = \; 16 \, n^3 \; + \; 48 \, n^2 \; + \; 44 \, n \; + \; 12$

then,

$a \, b \; + \; 1 \; = \; (n + 1)^2$
$a \, c \; + \; 1 \; = \; (2 \, n + 1)^2$
$a \, d \; + \; 1 \; = \; (4 \, n^2 + 6 \, n + 1)^2$
$b \, c \; + \; 1 \; = \; (2 \, n + 3)^2$
$b \, d \; + \; 1 \; = \; (4 \, n^2 + 10 \, n + 5)^2$
$c \, d \; + \; 1 \; = \; (8 \, n^2 + 16 \, n + 7)^2$

and

$((a + b) - (c + d))^2$
$= \; (n+n+2-4 \, n-4-16 \, n^3-48 \, n^2-44 \, n-12)^2$
$= \; (-16 \, n^3 - 48 \, n^2 - 46 \, n - 14)^2$
$= \; 4 \, (n + 1)^2 \; (8 \, n^2 + 16 \, n + 7)^2$

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

$((a + c) - (b + d))^2$
$= \; ((n+4 \, n+4)-(n+2+16 \, n^3+48 \, n^2+44 \, n+12))^2$
$= \; (-16 \, n^3 - 48 \, n^2 - 40 \, n - 10)^2$
$= \; 4 \, (2 \, n + 1)^2 \; (4 n^2 + 10 \, n + 5)^2$

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

$((a + d) - (b + c))^2$
$= \; ((n+16 \, n^3+48 \, n^2+44 \, n+12)-(n+2+4 \, n+4))^2$
$= \; (16 \, n^3 + 48 \, n^2 + 40 \, n + 6)^2$
$= \; 4 \, (2 \, n + 3)^2 \; (4 \, n^2 + 6 \, n + 1)^2$

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