1. Zadatak

Odmah vidimo da je $x>y$ pa mora biti $x-y>0$.

Jednadžbu možemo zapisati u obliku $(x-y)(x^2+xy+y^2)=91$.

Kako je $91=91\cdot1=7\cdot13$, imamo sljedeća četiri slučaja:

$$\begin{aligned} 1^\circ\quad&x-y=1,\quad x^2+xy+y^2=91\\ &y=x-1\\ &x^2+x(x-1)+(x-1)^2-91=0\\ &x^2-x-30=0\\ &x_1=6,\ x_2=-5\\ &y_1=5,\ y_2=-6. \end{aligned}$$ $$\begin{aligned} 2^\circ\quad&x-y=7,\quad x^2+xy+y^2=13\\ &y=x-7\\ &x^2+x(x-7)+(x-7)^2-13=0\\ &x^2-7x+12=0\\ &x_3=4,\ x_4=3\\ &y_3=-3,\ y_4=-4. \end{aligned}$$ $$\begin{aligned} 3^\circ\quad&x-y=91,\quad x^2+xy+y^2=1\\ &y=x-91\\ &x^2+x(x-91)+(x-91)^2-1=0\\ &x^2-91x+2760=0\\ &x_{5,6}=\frac{91\pm\sqrt{8281-11040}}2\notin\mathbb R. \end{aligned}$$ $$\begin{aligned} 4^\circ\quad&x-y=13,\quad x^2+xy+y^2=7\\ &y=x-13\\ &x^2+x(x-13)+(x-13)^2-7=0\\ &x^2-13x+54=0\\ &x_{7,8}=\frac{13\pm\sqrt{169-216}}2\notin\mathbb R. \end{aligned}$$

Dakle, sva cjelobrojna rješenja jednadžbe su: $(6,5)$, $(-5,-6)$, $(4,-3)$, $(3,-4)$.