Polynomial Conditions in 2015

Recall that, for all integers $a, b$ , we have $a - b | f(a) - f(b)$ . Thus, we expect that $20 - 15 | f(20) - f(15) = 15 - 9$ . This simplifies to $5|6$ , which is obviously not true. Thus, there exists no polynomial $f(x)$ with integer coordinates that satisfy the conditions, and the statement is $\boxed{\text{false}}$ .

Side note: A quick proof that $a - b | f(a) - f(b)$ . Let $f(x) = c_n x^n + c_{n - 1} x^{n - 1} + \cdots + c_1 x + c_0$ , where the $c_i$ are integers. Then,

$\begin{aligned} f(a) - f(b) &= c_n a^n + c_{n - 1} a^{n - 1} + \cdots + c_1 a + c_0 - (c_n b^n + c_{n - 1} b^{n - 1} + \cdots + c_1 b + c_0) \\ &= c_n(a^n - b^n) + c_{n - 1}(a^{n - 1} - b^{n - 1}) + \cdots + c_1(a - b) \\ &= (a - b)(\text{some expression}). \end{aligned}$

Thus, $a - b | f(a) - f(b)$ .

in modulo 5, f(15)=f(20)= $a_0$ ,the last term of the polynomial.But then f(20)=0 whereas f(15)=4 in modulo 5.contradiction.

I want to ask whether my solution is correct.

first case, let f(x) = ax

therefore, f(20)=15

thus, ax = 15

   a(20)=15 , therefore, 4a=3

also, f(15)=9, thus, similarly we have 5a=3

equating, we have, 3a=4a

thus no polynomial is possible like f(x)=ax

so consider f(x)=ax + bx

therefore, f(20)=15=a(20)+b(20) thus, 3=4a+4b

also, f(15)=9=a(15)+b(15) thus,3=5a+5b

equating we have, 5a+5b=4a+4b

thus no polynomial is possible like f(x)=ax+bx

hence similarly no polynomial is possible