+0

0
151
1

Feb 14, 2019

#1
+23073
+5

I can't give you a detailed solution, but i give you the solution by wolfram alpha:

WolframAlpha: solve  y^2-zx = -103, z^2-xy = 22, x^2-yz > 81 over the positive integers

So $$x^2-yz = 14^2-3\cdot 8 = 172$$

How do i get this $$x^2-yz > 81$$ ?

$$\begin{array}{|rcll|} \hline && (x^2-yz) + (y^2-zx) + (z^2-xy) \\ &=& x^2+y^2+z^2-yz-zx-xy \\\\ &=& \dfrac{1}{2}(2x^2+2y^2+2z^2-2yz-2zx-2xy) \\\\ &=& \dfrac{1}{2}\left( (x^2-2xy+y^2) + (y^2-2yz+z^2) + (z^2-2zx+x^2) \right) \\ &=& \dfrac{1}{2}\left( (x-y)^2 + (y-z)^2 + (z-x)^2 \right) \quad | \quad \text{this is always positive} \\\\ (x^2-yz) + (y^2-zx) + (z^2-xy) &>& 0 \\ (x^2-yz) -103 + 22 &>& 0 \\ (x^2-yz) -81 &>& 0 \\ \mathbf{x^2-yz} & \mathbf{>} & \mathbf{81} \\ \hline \end{array}$$

Feb 14, 2019
edited by heureka  Feb 14, 2019