## (a^2 + b^2)(c^2 + d^2)(e^2 + f^2) as the sum of two squares

We know that

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

so, applying this recursively,

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

if

$a \; = \; n$
$b \; = \; c \; = \; (n+1)$
$d \; = \; e \; = \; (n+2)$
$f \; = \; (n+3)$

then

$(n^2 \; + \; (n+1)^2) \, ((n+1)^2 \; + \; (n+2)^2) \, ((n+2)^2 \; + \; (n+3)^2)$
$= \; (2 \, n^3 + 8 \, n^2 + 9 \, n + 1)^2 \; + \; (2 \, n^3 + 10 \, n^2 + 15 \, n + 8)^2$
$= \; (2 \, n^3 + 8 \, n^2 + 11 \, n + 7)^2 \; + \; (2 \, n^3 + 10 \, n^2 + 13 \, n + 4)^2$
$= \; (2 \, n^3 + 8 \, n^2 + 7 \, n - 1)^2 \; + \; (2 \, n^3 + 10 \, n^2 + 17 \, n + 8)^2$
$= \; (2 \, n^3 + 12 \, n^2 + 19 \, n + 7)^2 \; + \; (2 \, n^3 + 6 \, n^2 + n - 4)^2$