Prove: There is no such polynomial function in $\mathbb{Z}[x]$ s.t. $f(7)=5$ and $f(15)=10$.
My first idea was to think $f(7)=5$ and $f(15)=10$ as points $(7,5)$ and $(15,10)$, but later on I couldn't go further.
The next idea is to construct two other functions $g(x)$ and $h(x)$ s.t. $g(x)=f(x)-5$, $h(x)=f(x)-10$. Then, $g(7)=0$, $h(15)=0$. Let $\displaystyle{f(x)=\sum_{k=0}^{n}a_kx^k}\;\;(\forall k\in\{0,1,2,\cdots,n\}, a_k\in \mathbb{Z}, a_n=1)$. By the rational root theorem, we have $$a_0-5\equiv0 \;\;(\operatorname{mod} 7)$$ $$a_0-10\equiv0\;\; (\operatorname{mod} 15).$$ Let $a_0=7a+5=15b+10\;(a,b\in\mathbb{Z})$. Then, $7a=15b+5$.
What am I supposed to do after this assumption? Am I going the right direction? Is there another way to interpret this problem?
