2021/069) Prove that there are no integers $a,\,b,\,c$ and $d$ such that the polynomial $ax^3+bx^2+cx+d$ equals 1 at $x=19$ and 2 at $x=62$.

We have  $f(62)−f(19)=a(62^3−19^3)+b(62^2−19^2)+c(62−19)=1 $

Or  $(62−19)(a(62^2+62∗19+19^2)+b(62+19)+c)=1$ 

LHS is a multiple of 43 and RHS is 1 so this does not have integer solution