A triple of positive integers $(a,b,c)$ is brazilian if $$a|bc+1$$$$b|ac+1$$$$c|ab+1$$Determine all the brazilian triples.
Problem
Source: Brazil National Olympiad Junior 2021 #8
Tags: number theory
16.02.2022 10:39
Easy to see that $anc|ab+bc+ca+1$ which means: $abcn=ab+bc+ca+1$ and know use inequalities yo prove that $RHS>LHS$ Similar too Greece nation Olympiad junior 2020
25.01.2023 07:48
Alright, here goes nothing. Multiplying the three relationships gives us, as P2nisic said, $abc \mid ab+bc+ca+1$ Which therefore means $abc \leq ab+bc+ca+1 \ (\star)$. If we assume $a \geq b \geq c$, then we have $ab+bc+ca + 1\leq 3a^2 + 1$, from which we conclude $abc \leq 3a^2 + 1 \implies a(bc-3a) \leq 1$. We have three cases: - $bc-3a = 1$ yields no solutions. - $bc-3a = 0$ also yields no solutions. Which leaves the case $bc < 3a$. From this inequality and $(\star)$ we conclude that: $abc \leq ab+bc+ca+1 < ab+3a+ca+1 = a(b+c+3)+1$ $\implies abc \leq a(b+c+3) \implies bc \leq b+c+3 \implies bc-b-c+1 = (b-1)(c-1) \leq 4$ We can divide this further into 5 cases, whose computations are fairly quick. Cases $(b-1)(c-1) = 4,3,1$ yield no solutions, while $(b-1)(c-1)=2$ yields $(7,3,2)$ and permutations as solutions, and $(b-1)(c-1) = 0$ yields $(3,2,1)$, $(2,1,1)$, $(1,1,1)$ and permutations as solutions.
25.01.2023 12:00