Решить задание во вложении

x^2 +y^2 -2z^2 =2a^2
z^2 -xy =a^2 lt;=gt; 2z^2 -2xy =2a^2
x^2 +y^2 -2z^2 = 2z^2 -2xy lt;=gt; x^2 +2xy +y^2 =4z^2 lt;=gt; (x+y)^2 = (2z)^2 lt;=gt; x+y=2z

[x+y=2z
[x+y= -2z

1)
x+y=2z
x+y+2z =4(a^2 +1)
2z+2z =4(a^2 +1) lt;=gt; 4z=4(a^2 +1) lt;=gt; z=a^2 +1 lt;=gt; z -a^2 =1

2)
x+y = -2z
-2z+2z = 4(a^2 +1)
4(a^2 +1)=0, но a^2 +1 gt;1.

x+y +z -3a^2 = 2z +z -3a^2 = 3(z -a^2) = 3