Functional Equation

Bài toán : Tìm hàm số $f:{\mathbb{N}}^{\ast }\to {\mathbb{N}}^{\ast }$ và thỏa mãn :

$$\{\begin{array}{c}f(mn)=f(m)+f(n)\\ f(n)=0,\forall n\equiv 3(mod10)\\ f(10)=0\end{array}$$

Lời giải :

Ta có $0=f(10)=f(1)+f(10)=f(2)+f(5)\Rightarrow f(1)=f(2)=f(5)=0$

Ta có $f(3)=0\Rightarrow f(6)=f(2)+f(3)=0$

$$f(4)=f(2)+f(2)=0,f(8)=f(4)+f(4)=0$$

$$f(9)=f(3)+f(3)=0,f(63)=0\Rightarrow f(7)=f(63)-f(9)=0$$

Như vậy ta có $f(1)=f(2)=...=f(10)=0$

Ta giả sử $f(n)=0,\forall n\in \mathbb{N},n=\overline{1,k}$. Xét với $k+1$ :

• Nếu $k+1\equiv 0,2,4,6,8(mod10)$ thì :

$$f(k+1)=f\left(2.\frac{k+1}{2}\right)=f(2)+f\left(\frac{k+1}{2}\right)=0$$

• Nếu $k+1\equiv 1(mod10)\Rightarrow 3(k+1)\equiv 3(mod10)$

$$f(3(k+1))=f(3)+f(k+1)\Rightarrow f(k+1)=f(3(k+1))-f(3)=0$$

• Nếu $k+1\equiv 3(mod10)\Rightarrow f(k+1)=0$
• Nếu $k+1\equiv 5(mod10)\Rightarrow f(k+1)=f\left(5\right)+f\left(\frac{k+1}{5}\right)=0$
• Nếu $k+1\equiv 7(mod10)\Rightarrow 9(k+1)\equiv 3(mod10)\Rightarrow f(k+1)=f(9(k+1))-f(9)=0$
• Nếu $k+1\equiv 9(mod10)\Rightarrow 7(k+1)\equiv 3(mod10)\Rightarrow f(k+1)=f(7(k+1))-f(7)=0$

Theo nguyên lí quy nạp ta có $f(n)=0,\forall n\in {\mathbb{N}}^{\ast }$