## Oblong numbers : x(x+ 1), y(y+ 1), z(z+ 1) in arithmetic progression

An Oblong number is a number which is the product of two consecutive integers, that is, a number of the form   $n \,(n + 1)$

There exist infinitely many triplets of positive integers   $x, \; y, \; z$,   for which the numbers
$x \,(x+1)$,     $y \,(y+1)$,     $z \,(z+1)$
form an increasing arithmetic progression.

The required property holds for

(1)

$x \; = \; n$
$y \; = \; 5 \, n \; + \; 2$
$z \; = \; 7 \, n \; + \; 3$

n   is a positive integer

since in this case the numbers

$n \,(n+1)$,     $(5 \, n+2) \,(5 \, n+3)$,     $(7 \, n+3) \,(7 \, n+4)$

form the arithmetic progression with the common difference   $6 \, (2 \, n+1)^2$

$(5 \, n+2) \,(5 \, n+3) \; - \; n \,(n+1)$
$= \; (25 \, n^2 + 25 \, n + 6) \; - \; (n^2 + n)$
$= \; 24 \, n^2 \; + \; 24 \, n \; + \; 6$
$= \; 6 \, (2 \, n+1)^2$

$(7 \, n+3) \,(7 \, n+4) \; - \; (5 \, n+2) \,(5 \, n+3)$
$= \; (49 \, n^2 + 49 \, n + 12) \; - \; (25 \, n^2 + 25 \, n + 6)$
$= \; 24 \, n^2 \; + \; 24 \, n \; + \; 6$
$= \; 6 \, (2 \, n + 1)^2$

and for

(2)

the numbers

$x \; = \; n$,
$y \; = \; 29 \,n \; + \; 14$,
$z \; = \; 41 \,n \; + \; 20$

form an arithmetic progression:

$(29 \, n+14) \,(29 \, n+15) \; - \; n \,(n+1) \; = \; 210 \, (2 \, n + 1)^2$
$(41 \, n+20) \,(41*n+21) \; - \; (29 \, n+14) \,(29 \, n+15) \; = \; 210 \, (2 \, n + 1)^2$

$(29 \, n+14)(29 \, n+15) - n \,(n+1) = (41 \, n+20)(41 \, n+21) - (29 \, n+14)(29 \, n+15)$