Macedonian Mathematical Olympiad

After dividing the equation by $xyzt$ we get $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}=1+\frac{3}{xyzt}$. Because of symmetry, without loss of generality, we can assume that $$x\le y\le z\le t\dots \text{\hspace{1em}(1)}$$ from where it follows that $\frac{1}{x}\ge \frac{1}{y}\ge \frac{1}{z}\ge \frac{1}{t}$. We get $\frac{4}{x}\ge \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}=1+\frac{3}{xyzt}>1$, from where we have $x<4$.

Case 1. Let $x=3$. Then the equation is of the form $3yz+yzt+3zt+3yt=3yzt+3$, or, equivalently $3(yz+zt+yt)=2yzt+3$. After dividing this equation by $yzt$ we get $$3\left(\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right)=2+\frac{3}{yzt}>2,\frac{9}{y}>2,\text{ from where we have }y\le 4.$$ The possible values for $y$ are 3 and 4.

a) For $y=4$ we get $$3(4z+zt+4t)=8zt+3,12(z+t)=5zt+3,12\left(\frac{1}{z}+\frac{1}{t}\right)=5+\frac{3}{zt}>5,\frac{24}{z}>5,$$ from where we have $z\le 4$. From (1) it follows that $z=4$ and the equation gets the form $12(4+t)=20t+3$, or, equivalently $8t=45$, which implies that $t$ is not a natural number.

b) For $y=3$, we get $$3(3z+zt+3t)=6zt+3,3(z+t)=zt+1,3\left(\frac{1}{z}+\frac{1}{t}\right)=1+\frac{1}{zt}>1,\frac{6}{z}>1,z<6.$$ The possible values for $z$ are 3, 4, 5. - Let $z=3$. Then $3(3+t)=3t+1$ which is impossible. - If $z=4$, then $3(4+t)=4t+1$, $t=11$. - If $z=5$, then $3(5+t)=5t+1$, $t=7$. We get that the quadriplets $(3,3,4,11)$, $(3,3,5,7)$ are solutions.

Case 2. Let $x=2$. Then the equation is of the form $$2yz+yzt+2zt+2yt=2yzt+3,$$ or, equivalently, $$2(yz+zt+yt)=yzt+3(2).$$ Then each of the numbers $y,z,t$ is odd. After dividing this equation by $yzt$ we get $2\left(\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right)=1+\frac{3}{yzt}>1$ from where we have $\frac{6}{y}>1$, or, equivalently $y<6$.

a) If $y=5$ then (2) is of the form $2(5z+zt+5t)=5zt+3$, or, equivalently $10(z+t)=3zt+3$. Hence $10\left(\frac{1}{z}+\frac{1}{t}\right)=3+\frac{3}{zt}>3$, therefore $\frac{1}{z}>\frac{3}{20}$, or equivalently $z\le 6$. The only possibility is $z=5$. We get $10(5+t)=15t+3$, or, equivalently $5t=47$ which implies that $t$ is not a natural number.

b) If $y=3$, (2) is of the form $2(3z+zt+3t)=3zt+3$, or equivalently $6(z+t)=zt+3$. Then $6\left(\frac{1}{z}+\frac{1}{t}\right)=1+\frac{3}{zt}>1$, from where $\frac{12}{z}>1$, or, equivalently $z<12$. The possibilities for $z$ are 3, 5, 7, 9, 11. - If $z=3$, then $6(3+t)=3t+3$, from where we have $3t=-15$, or equivalently $t=-5\notin \mathbb{N}$. - If $z=5$, then $6(5+t)=5t+3$, $t=-27\notin \mathbb{N}$. - If $z=7$, then $6(7+t)=7t+3$, $t=39$. - If $z=9$, then $6(9+t)=9t+3$, from where we have $3t=51$, or, equivalently $t=17$. Therefore in this case the solutions are the quadriplets $(2,3,7,39)$, $(2,3,9,17)$.

Case 3. The case remains when $x=1$. Then the equation is of the form $yz+yzt+zt+yt=yzt+3$, or, equivalently $yz+zt+yt=3$. From (1) we get $3yz\le 3$, or equivalently $yz\le 1$, from where $y=1$ and $z=1$. Then $1+2t=3$, or equivalently $t=1$. The quadriple $(1,1,1,1)$ is a solution.

Finally, the solutions to the initial equation are all permutations of $(3,3,4,11)$, $(3,3,5,7)$, $(2,3,7,39)$, $(2,3,9,17)$, $(1,1,1,1)$.