65th Czech and Slovak Mathematical Olympiad

Taking the product of the 1st, the 3rd and the 5th fractions reveals that their value has to be 1, that is $$a+b=c+d=e+f=p.\text{\hspace{1em}(1)}$$ the form of the second and of the fourth fraction implies $$f+a\mid d+e\text{and}d+e\mid b+c.\text{\hspace{1em}(2)}$$ that is first $f+a$ is at most the arithmetic mean of its multiples, $$f+a\le \frac{1}{3}((f+a)+(d+e)+(b+c))=p,\text{\hspace{1em}(3)}$$ and $$f+a\mid (f+a)+(d+e)+(b+c)=3p.$$ Thus $f+a$ divides $3p$ and is in the interval $⟨2,p⟩$. Consequently either $f+a=p$ or $f+a=3$. We deal separately with these cases.

i) Let $f+a=p$. Because of (3) there is $f+a=d+e=b+c=p$, which together with (1) gives $p-1$ solutions of the form $$(a,b,c,d,e,f)=(a,p-a,a,p-a,a,p-a),\text{where }a\in \{1,2,\dots ,p-1\}.$$

ii) Let $f+a=3$. Then $\{a,f\}=\{1,2\}$.

Firstly let $a=1$ and $f=2$. According to (1) then $b=p-1$ and $e=p-2$, and (2) has the form $$3\mid d+(p-2)\text{and}d+(p-2)\mid (p-1)+c.\text{\hspace{1em}(4)}$$ In analyzing (4) we distinguish between $d=1$ and $d\ge 2$.

If $d=1$ then $c=p-1$ and (4) reads $$3\mid p-1\text{a}p-1\mid 2(p-1).$$ While the right relation always holds, the left one holds only for $p=3q+1$ ($q$ is a suitable positive integer). For such prime numbers we get considering (1) solutions $$(a,b,c,d,e,f)=(1,p-1,p-1,1,p-2,2).$$ If $d\ge 2$ we show first, that the right relation in (4) is satisfied if and only if $d+(p-2)=(p-1)+c$ or $d=c+1$. $d\ge 2$ namely implies $c=p-d\le p-2$, thus $$d+(p-2)\ge p\text{and}(p-1)+c\le 2p-3<2p$$ and $d+(p-2)=(p-1)+c$. From $c+d=p$ and $d=c+1$ we get $c=\frac{1}{2}(p-1)$ a $d=\frac{1}{2}(p+1)$. Since $d+(p-2)=\frac{3}{2}(p-1)$, the left relation in (4) is fulfilled and $$(a,b,c,d,e,f)=(1,p-1,\frac{1}{2}(p-1),\frac{1}{2}(p+1),p-2,2).$$ is a solution.

Finally $a=2$ and $f=1$. In this case $b=p-2$ a $e=p-1$, and (2) reads $$3\mid d+(p-1)\text{and}d+(p-1)\mid (p-2)+c.(5)$$ Because $$d+(p-1)\ge p\text{and}(p-2)+c<2p,$$ the right relation in (5) holds if and only if $d+(p-1)=(p-2)+c$, that is iff $c=d+1$. Together with $c+d=p$ we get $c=\frac{1}{2}(p+1)$ and $d=\frac{1}{2}(p-1)$, thus the right relation in (5) holds as well, and the last solution is $$(a,b,c,d,e,f)=(2,p-2,\frac{1}{2}(p+1),\frac{1}{2}(p-1),p-1,1).$$

Conclusion. All the solutions found are apparently mutually different and their number depends on $p$ modulo $3$ ($p>3$): If $p=3q+1$ then there are $p+2$ sextuples, if $p=3q+2$, there are $p+1$ sextuples.