Câu 3: Hãy xác định tất cả các bộ nguyên dương (a,b) sao cho $a^2.b+a+b$ chia hết cho $a.b^2+b+7$
Lời giải. Chỉ có hai trường hợp sau xảy ra:
$\fbox{TH1}.$Nếu $a<b$, khi đó $b \ge a+1$. Từ đó $$ab^2+b+7>ab^2+b=(ab+1)b \ge (a+1)(b+1),$$
suy ra $$ab^2+b+7>a^2b+a+ab+1>a^2b+a+b$$
Như vậy thì $$(ab^2+b+7) \nmid (a^2b+b+7) $$
Mâu thuẫn với giả thiết bài toán.
$\fbox{TH2}.$ Nếu $a \ge b.$ Đặt $k= \frac{a^2+a+b}{ab^2+b+7}$.
Ta có $$\left( \frac{1}{b}+ \frac{a}{b} \right) (ab^2+b+7)=a^2b+a+ab+7. \frac{a}{b}+ \frac{7}{b}+1>a^2b+a+b$$
Suy ra $$\frac{a}{b}+ \frac{1}{b}> \frac{a^2b+a+b}{ab^2+b+7}=k.$$
Có ba khả năng sau xảy ra:
- Nếu $b \ge 3 \Rightarrow \left( b- \frac{7}{b} \right) > 0.$ Từ đó ta có
$$\left( \frac{a}{b}- \frac{1}{b} \right) (ab^2+b+7)=a^2b+a-a \left(b- \frac{7}{b} \right)-1- \frac{7}{b}<a^2b+a<a^2b+a+b$$
Suy ra $$\begin{aligned}\frac{a}{b}- \frac{1}{b} & < \frac{a^2b+a+b}{ab^2+b+7}=k \\ & \Rightarrow \frac{a-1}{b}<k< \frac{a+1}{b} \\ & \Rightarrow a-1<kb<a+1. \end{aligned}$$
Như vậy $a=kb$, thay $k= \frac{a^2+a+b}{ab^2+b+7}$ ta có
$$ \begin{aligned} ab^2k+bk+7k & =a^2b+a+b \\ & \Rightarrow k^2b^3+kb+7k=k^2b^3+kb+b \\ & \Rightarrow b=7k \\ & \Rightarrow a=7k^2 \end{aligned}$$
Trong trường hợp này thì $\boxed{(a,b)=(7k^2,7k)}$ với $k$ nguyên dương.
- Nếu $b=1$. Ta tìm được nghiệm $\boxed{(a,b) \in \{ (11,1),(49,1) \}. }$
- Nếu $b=2$ thì phương trình vô nghiệm.
Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.
Grothendieck, Récoltes et Semailles (“Crops and Seeds”).