Solve in rational numbers the system of equations — Part 2

$x \; + \; y \; + \; z \; + \; x^2 \; = \; a^2$

$x \; + \; y \; + \; z \; + \; x^2 \; + \; y^2 \; = \; b^2$

$x \; + \; y \; + \; z \; + \; x^2 \; + \; y^2 \; + \; z^2 \; = \; c^2$

Solutions:

(1)

Integral solutions:     $x = 35$,   $y = 144$,   $z = 360$

$35 + 144 + 360 + 35^2 = 1764 = 42^2$

$35 + 144 + 360 + 35^2 + 144^2 = 22500 = 150^2$

$35 + 144 + 360 + 35^2 + 144^2 + 360^2 = 152100 = 390^2$

(2)

$x = 16/5$,     $y = 48/5$,     $z = 144/5$

$(16/5) + (48/5) + (144/5) + (16/5)^2 = (36/5)^2$

$(16/5) + (48/5) + (144/5) + (16/5)^2 + (48/5)^2 = 12^2$

$(16/5) + (48/5) + (144/5) + (16/5)^2 + (48/5)^2 + (144/5)^2 = (156/5)^2$

(3)

Not only can three numbers be found by this method but also four can be found by means of four square numbers:

$x = 1295$,     $y = 31968/7$,     $z = 79920/7$,     $w = 79920$

$x + y + z + w + x^2$

$1295 + 31968/7 + 79920/7 + 79920 + 1295^2 = 1774224 = 1332^2$

$x + y + z + w + x^2 + y^2$

$1295 + (31968/7) + (79920/7) + 79920 + 1295^2 + (31968/7)^2$
$= 1108890000/49$
$= ( \,33300/7 \,)^2$

$x + y + z + w + x^2 + y^2 + z^2$

$1295 + (31968/7) + (79920/7) + 79920 + 1295^2 + (31968/7)^2 + (79920/7)^2$
$= 7496096400/49$
$= ( \,86580/7 \,)^2$

$x + y + z + w + x^2 + y^2 + z^2 + w^2$

$1295 + (31968/7) + (79920/7) + 79920 + 1295^2 + (31968/7)^2 + (79920/7)^2 + 79920^2$
$= 320469210000/49$
$= ( \,566100/7 \,)^2$