Balkan Mathematical Olympiad

Given equation is equivalent to $abc=p(ab+bc+ca)$, and therefore $p\mid abc$. So, at least one of $a,b,c$, w.l.o.g. $a$, is divisible by $p$ and let $a=p{a}_1$. First, let us suppose that $p\nmid bc$. Then $$\frac{b+c}{bc}=\frac{1}{b}+\frac{1}{c}=\frac{1}{p}-\frac{1}{a}=\frac{a_1-1}{p{a}_1},$$ i.e. $p{a}_1(b+c)=bc(a_1-1)$, and therefore $p\mid a_1-1$. Since $a_1>1$ we have $a_1\ge p+1$, and therefore $a+b+c>p{a}_1\ge p(p+1)\ge 2p\sqrt{p}$, a contradiction.

Next, let us assume that $p\mid b$, but $p\nmid c$. Let $b=p{b}_1$. Then $$\frac{1}{c}=\frac{1}{p}-\frac{1}{a}-\frac{1}{b}=\frac{a_1{b}_1-a_1-b_1}{p{a}_1{b}_1},$$ i.e. $p{a}_1{b}_1=c(a_1{b}_1-a_1-b_1)$, and therefore $p\mid a_1{b}_1-a_1-b_1$. If $a_1{b}_1-a_1-b_1=0$, then $(a_1-1)(b_1-1)=1$, and therefore $a_1=b_1=2$. But then $\frac{1}{a}+\frac{1}{b}=\frac{1}{p}$, a contradiction. So, $a_1{b}_1-a_1-b_1\ge p$, and therefore $a_1{b}_1>p$. By AG inequality $a_1+b_1\ge 2\sqrt{a_1{b}_1}>2\sqrt{p}$, and therefore $$a+b+c>p(a_1+b_1)>2p\sqrt{p},$$ a contradiction.

So, each of the numbers $a,b,c$ is divisible by $p$. Let $a=a_{1}p,b=p{b}_1,c=p{c}_1$. Given equation is equivalent to $$\frac{1}{a_1}+\frac{1}{b_1}+\frac{1}{c_1}=1.$$ Let $m=\min \{a_1,b_1,c_1\}$. Then $$1=\frac{1}{a_1}+\frac{1}{b_1}+\frac{1}{c_1}\le \frac{3}{m},$$ and therefore $m\le 3$.

* If $m=3$, then $a_1=b_1=c_1=3$, and $a=b=c=3p$. This is a solution of the problem if and only if $a+b+c=9p<2p\sqrt{p}$, i.e. if and only if $p\ge 23$.

* If $m=2$, and w.l.o.g. $a_1=2$, then $\frac{1}{b_1}+\frac{1}{c_1}=\frac{1}{2}$. Let $m^{\prime }=\min \{b_1,c_1\}$. As in the previous part of the solution we get that $m^{\prime }\le 4$. If $m^{\prime }=4$, then $b_1=c_1=4$, i.e. $a=2p,b=c=4p$. This is a solution of the problem if and only if $a+b+c=10p<2p\sqrt{p}$, i.e. $p\ge 29$. If $m^{\prime }=3$, then $\{b_1,c_1\}=\{3,6\}$. This is a solution of the problem if and only if $a+b+c=11p<2p\sqrt{p}$, i.e. if and only if $p\ge 31$.

All solutions of the problem are permutation of the triples: $(3p,3p,3p),(2p,4p,4p),(2p,3p,6p)$, for $p\ge 31$ $(3p,3p,3p),(2p,4p,4p)$, for $p=29$ * $(3p,3p,3p)$, for $p=23$.