Question

[FFPUBGAOVCFLOL]

Cho x,a,b \(\in\)N* thỏa mãn \(x+3=2^a\)và \(3x+1=4^b\). Tìm x,a,b.

Answer 1

[Kiệt Nguyễn]

Vì \(x\inℕ^∗\)nên \(x\ge1\)

\(\Rightarrow2x\ge2\Leftrightarrow3x+1\ge x+3\)

\(\Rightarrow4^b>2^a\Rightarrow4^b⋮2^a\)

\(\Rightarrow3x+1⋮x+3\)

\(\Rightarrow3\left(x+3\right)-8⋮x+3\)

Mà \(3\left(x+3\right)⋮x+3\)nên \(8⋮x+3\)

\(\Rightarrow x+3\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

Mà \(x+3\ge4\)(do x nguyên dương) nên \(x+3\in\left\{4;8\right\}\)

+) Xét x + 3 = 4 \(\Leftrightarrow2^a=4\Leftrightarrow a=2\)

\(x+3=4\Rightarrow x=1\Rightarrow3x+1=4=4^b\Rightarrow b=1\)

+) Xét x + 3 = 8 \(\Leftrightarrow2^a=8\Leftrightarrow a=3\)

\(x+3=8\Rightarrow x=5\Rightarrow3x+1=16=4^b\Rightarrow b=2\)

Vậy ta tìm được bộ ba số (a;b;x) thỏa mãn là \(\left(2;1;1\right);\left(3;2;5\right)\)