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:

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