$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{x^2+xy+y^2}$ fine $x,y$ :
$$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{x^2+xy+y^2} \ \ \ : x ,y \in \mathbb{Z}$$
My Try :
$$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{x^2+xy+y^2} \\ x^2-x+1+y^2-y+1+2\sqrt{(x^2-x+1)(y^2-y+1)}=x^2+xy+y^2 \\ 2\sqrt{(x^2-x+1)(y^2-y+1)}=xy +x+y-2$$
Now ?

Applying Minkowski's inequality, we have that $$ \sqrt{x^{2}-x+1}+\sqrt{y^{2}-y+1}=\big((x-\frac{1}{2})^{2}+\big(\frac{\sqrt{3}}{2}\big)\big)^{\frac{1}{2}}+\big((y-\frac{1}{2})^{2}+\big(\frac{\sqrt{3}}{2}\big)\big)^{\frac{1}{2}} \geq \big(\big(x-\frac{1}{2}+y-\frac{1}{2}\big)^{2}+(\sqrt{3})^{2}\big)^{\frac{1}{2}}=\sqrt{x^{2}+y^{2}+2xy-2x-2y+4} $$
Therefore, using the hypothesis, we get that $$ \sqrt{x^{2}+xy+y^{2}} \geq \sqrt{x^{2}+y^{2}+2xy-2x-2y+4} $$
which after squaring and some computation reduces to $ (x-2)(y-2) \le 0 $.
We apply Minkowsi again but this time in the form $$ \sqrt{x^{2}-x+1}+\sqrt{y^{2}-y+1}=\big(\big(x-\frac{1}{2})^{2}+\big(\frac{\sqrt{3}}{2}\big)^{2}\big)^{\frac{1}{2}}+\big(\big(\frac{1}{2}-y)^{2}+\big(\frac{\sqrt{3}}{2}\big)^{2}\big)^{\frac{1}{2}} \ge \big((x-y)^{2}+(\sqrt{3})^{2}\big)^{\frac{1}{2}} $$
Hence $$ \sqrt{x^{2}+xy+y^{2}} \geq \sqrt{x^{2}+y^{2}-2xy+3} $$
Squaring again, we get that $ xy \ge 1 $. 
The condition $ (x-2)(y-2) \le 0 $ together with $ xy \ge 1 $ tell us that one of $ x $ and $ y $ must necessarily be $ 1 $ or $ 2 $. WLOG, suppose $ x \in \{1,2\} $. We plug in the value of $ x $ in the original equation and computation then shows that the only possible solution is $ x=y=2 $ and this indeed satisfies the initial equation.
