Just a question

We have $p(x)=ax^{3}+bx^{2}+cx+d$ a third degree polynomial. From the conditions $p(15)=9$ and $p(7)=5$ we get simultaneously:

$a15^{3}+b15^{2}+c15+d=9$ and $a7^{3}+b7^{2}+c7+d=5$ .

By substracting the second equation from the first equation we get: $a(15^{3}-7^{3})+b(15^{2}-7^{2})+c(15-7)=4 \Rightarrow a(15-7)(15^{2}+105+7^{2})+b(15-7)(15+7)+c(15-7)=4 \Rightarrow 8[a(225+105+49)+b22+c]=4$ and by dividing by $8$ ,we get $379a+22b+c=\frac{1}{2}$ . Given that the coefficients of $a,b$ and $c$ are integer numbers and $\frac{1}{2}$ is a rational number $\Rightarrow$ there can be no integer coefficients $a,b$ and $c$ such as to respect the given conditions. Therefore, the answer is None .

Just use the fact that for a polynomial $p(x)$ with integer coefficients and integers $a,b$ , we have $a-b|p(a)-p(b)$ . Hence, in this case, we must have $15-7 | p(15)-p(7)$ , i.e., $8|4$ , which is impossible. Hence, there is no such polynomial.