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Belarus — Selection and Training Session

Algebra Difficulty 4.9 AIME Prove it Belarus

Do there exist six pairwise distinct positive integers $a$ , $b$ , $c$ , $d$ , $e$ and $m$ , such that
$\begin{cases} a+b+c = d+e+m, \\ ab+bc+ac = de+em+dm, \\ abc = dem + 3^{2021} \cdot 2^{2022}? \end{cases}$
(Igor Voronovich)

Solution

Answer: yes, such numbers exist.
Consider the triplets $(a, b, c) = (13, 4, 3)$ and $(d, e, m) = (12, 7, 1)$ . These numbers satisfy the system
$\begin{cases} a+b+c = 20 = d+e+m, \\ ab+bc+ac = 103 = de+em+dm, \\ abc - dem = 3^2 \cdot 2^3. \end{cases}$
Clearly, for $k = 2^{673} \cdot 3^{673}$ the numbers $ak$ , $bk$ , $ck$ , $dk$ , $ek$ and $mk$ satisfy the problem conditions.

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