I need to solve this problem and I don’t know how.
If $y^2 + z^2 + yz = a^2$
$z^2 + x^2 + zx = b^2$
$x^2 + y^2 + xy = c^2$
$yz + zx + xy = 0,$ then
$a \pm b \pm c = 0$
I can see that $a^2 + b^2 + c^2 = 2(x^2 + y^2 + z^2)$ which I suppose can be useful but then I don’t know what to do next. Can you give me a hint ? Thanks.
Because $$a^2+b^2+c^2=\sum_{cyc}(2x^2+xy)=\sum_{cyc}(2x^2+4xy)=2(x+y+z)^2$$ and $$\sum_{cyc}a^2b^2=\sum_{cyc}(x^2+xy+y^2)(x^2+xz+z^2)=$$ $$=(x+y+z)^4-3(x+y+z)^2(xy+xz+yz)+3(xy+xz+yz)^2=(x+y+z)^4.$$ Id est, $$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=\sum_{cyc}(2a^2b^2-a^4)=$$ $$=4\sum_{cyc}a^2b^2-(a^2+b^2+c^2)^2=4(x+y+z)^4-4(x+y+z)^4=0$$ and we are done!
The cyclic sum works so.
For three variables $a$, $b$ and $c$ we have: $$\sum_{cyc}a=a+b+c,$$ $$\sum_{cyc}ab=ab+bc+ca,$$ $$\sum_{cyc}a^2b=a^2b+b^2c+c^2a,$$ $$\sum_{cyc}a^2b^2=a^2b^2+b^2c^2+c^2a^2$$ and something interesting: $$(a-b)^2(a-c)^2(b-c)^2=$$ $$=\sum_{cyc}(a^4b^2+a^4c^2-2a^3b^3-2a^4bc+2a^3b^2c+2a^3c^2b-2a^2b^2c^2).$$