Đặt \(f\left(x\right)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0\)\(\left(a_i\in Z\right)\)
Ta có: \(f\left(15\right)=a_n.15^n+a_{n-1}.15^{n-1}+...+a_1.15+a_0=9\)
\(f\left(7\right)=a_n.7^n+...+a_1.7+a_0=5\)
\(\Rightarrow\left(15^n-7^n\right)a_n+\left(15^{n-1}-7^{n-1}\right).a_{n-1}+...+\left(15-7\right)a_1=9-5\)
Mà \(15^k-7^k=\left(15-7\right)\left(15^{k-1}+15^{k-2}.7+...+15^i.7^{k-1-i}+..+15.7^{k-2}+7^{k-1}\right)=8X_k\)
\(\left(X_K\in Z\right)\)
\(\Rightarrow8X_n.a_n+8X_{n-1}.a_{n-1}+...+8a_1=4\)
\(\Rightarrow X_na_n+X_{n-1}a_{n-1}+...+X_1a_1=\frac{1}{2}\text{ (vô lí do }X_k,\text{ }a_k\in Z\text{)}\)
Vậy không tồn tại đa thức hệ số nguyên thỏa f(7) = 5; f(15) = 9.
