6. Zadatak

Rastavimo lijevu stranu jednadžbe na faktore.

$$\begin{aligned} xyz+xy+xz+yz+x+y+z+1 &=(xyz+xy)+(xz+x)+(yz+y)+(z+1)\\ &=xy(z+1)+x(z+1)+y(z+1)+(z+1)\\ &=(xy+x+y+1)(z+1)\\ &=(x(y+1)+(y+1))(z+1)\\ &=(x+1)(y+1)(z+1). \end{aligned}$$

Kako je $2020=1\cdot2^2\cdot5\cdot101$ i vrijedi

$$2\leq x+1<y+1<z+1,$$

dobivamo tri mogućnosti:

$$\begin{array}{llll} \text{(i)}&x+1=2,&y+1=5,&z+1=202,\\ \text{(ii)}&x+1=2,&y+1=10,&z+1=101,\\ \text{(iii)}&x+1=4,&y+1=5,&z+1=101, \end{array}$$

odakle dobivamo tražena rješenja

$$(x,y,z)\in\{(1,4,201),(1,9,100),(3,4,100)\}.$$