Tìm x, y, z nguyên dương biết xyz = x + y + z + 9

Lời giải:

Không mất tính tổng quát ta giả sử 1 ≤ x ≤ y ≤ z

Chia cả 2 vế cho xyz > 0 ta được:

\(\frac{1}{{yz}} + \frac{1}{{xz}} + \frac{1}{{xy}} + \frac{9}{{xyz}} = 1\)

Vì 1 ≤ x ≤ y ≤ z nên x² ≤ xy ≤ xz ≤ yz ≤ xyz

Suy ra: \[1 = \frac{1}{{yz}} + \frac{1}{{xz}} + \frac{1}{{xy}} + \frac{9}{{xyz}} \le \frac{{12}}{{{x^2}}}\]

Mà x nguyên nên x ∈ {1; 2; 3}

+ Với x = 1 thì \(\frac{1}{{yz}} + \frac{1}{z} + \frac{1}{y} + \frac{9}{{yz}} = 1\)

⇔ 1 + y + z + 9 = yz

⇔ (z – 1)(y – 1) = 11

Suy ra: \(\left[ \begin{array}{l}\left\{ \begin{array}{l}z - 1 = 1\\y - 1 = 11\end{array} \right.\\\left\{ \begin{array}{l}z - 1 = 11\\y - 1 = 1\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}z = 2\\y = 12\end{array} \right.\\\left\{ \begin{array}{l}z = 12\\y = 2\end{array} \right.\end{array} \right.\)

⇒ (x;y;z) ∈ {(1;12;2), (1;2;12)}

+ Với x = 2 thì \(\frac{1}{{yz}} + \frac{1}{{2z}} + \frac{1}{{2y}} + \frac{9}{{2yz}} = 1\)

⇔ y + z – 2yz + 11 = 0

⇔ 2y + 2z – 4yz + 22 = 0

⇔ (2z – 1)(2y – 1) = 23

Suy ra: \(\left\{ \begin{array}{l}2z - 1 = 1\\2y - 1 = 23\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}z = 1\\y = 12\end{array} \right.\)

⇒ (x;y;z) ∈ {(2;12;1), (2;1;12)}

+ Với x = 3 thì \(\frac{1}{{yz}} + \frac{1}{{3z}} + \frac{1}{{3y}} + \frac{9}{{3yz}} = 1\)

⇔ y + z + 3 + 9 = 3yz

⇔ 3y + 3z – 9yx + 36 = 0

⇔ (3z – 1)(3y – 1) = 37

Suy ra: \(\left\{ \begin{array}{l}3z - 1 = 1\\3y - 1 = 37\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}z = \frac{2}{3} \notin \mathbb{Z}\\y = \frac{{38}}{3} \notin \mathbb{Z}\end{array} \right.\)

Vậy (x;y;z) ∈ {(1;12;2), (1;2;12); (2;12;1), (2;1;12)}