System of equations... Can YOU find $n$ ?

By squaring the first equation, you get $x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = 0$ From the second equation, It can be substituted in to give $\sqrt{74} + 2xy + 2yz + 2zx = 0$

Similarly, squaring the second equation gives $x^4 + y^4 + z^4 + 2x^2 y^2 + 2y^2 z^2 + 2z^2 x^2 = 74$ which when substituted with the third equation gives $n+ 2x^2 y^2 + 2y^2 z^2 + 2z^2 x^2 = 74$ . To get rid of the duel terms, I square the equation from earlier to give the another equation $(2xy + 2yz + 2zx)^2 = 74\\$ $\implies x^2 y^2 + y^2 z^2 + z^2 x^2 + 2x^2 yz + 2xy^2 z + 2xyz^2 = 74/4\\$ $\implies x^2 y^2 + y^2 z^2 + z^2 x^2 + xyz(x+y+z) = 74/4$ Using the first equation tells us that x + y + z = 0 so we can ignore the xyz(x+y+z) term because it is zero. Hence $x^2 y^2 + y^2 z^2 + z^2 x^2 = 74/4$

Using $x^4 + y^4 + z^4 + 2x^2 y^2 + 2y^2 z^2 + 2z^2 x^2 = 74$ from before, This now gives $n + 2(74/4) = 74\\$ $n = 74/2 = 37\\$ $n - 7 = 30\\$ $\boxed{\frac{n-7}{2} = 15}$