Tìm tất cả các số nguyên dương $a,b,c,d$ thỏa mãn:
$2^{a}=3^{b}-5^{c}+7^{d}$
ủa nếu là nguyên dương thì vô nghiệm ngay r nhỉ :3 –  tran85295 03-06-16 08:21 PM
Nhận thấy $3^b-5^c+7^d$ chỉ có thể là số lẻ
$\Rightarrow 2^a \hspace{1mm} \text{lẻ}\Leftrightarrow a=0$ 
Phương trình trở thành $3^b+7^d=5^c+1 \quad (*)$
Với $c=0$ thì $b=d=0$ tương tự xét với $b=0,d=0$
Với $b.c.d \ne0$
Bằng pp quy nạp dễ chứng minh $7^d \equiv 1(\mod3) $
$\Rightarrow 7^d+3^b \equiv 1(\mod{3}) \quad (1)$
Ta lại có $\left[ \begin{array}{l} 5^c \equiv 1(\mod 3)\\ 5^c \equiv 2(\mod 3) \end{array} \right.\Leftrightarrow \left[ \begin{array}{l}  5^c+1 \equiv 2 (\mod 3)\\ 5^c+1 \equiv0 (\mod3) \end{array} \right. \quad (2)$
Từ $(1),(2)\Rightarrow (*)$ vô nghiệm với $b.c.d \ne0$
Vậy  ta có nghiệm duy nhất $a=b=c=d=0$

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