# Cubic Polynomial

• MHB
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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 $x=62$.

HOI
Well, to start with, since $ax^3+ bx^2+ cx+ d$ is 1 when x 19, $a(19)^3+ b(19)^2+ 19x+ d= 6859a+ 361b+ 19c+ d= 1$, And since it is 2 when x= 62, $a(62)^3+ b(62)^2+ a(62)+ d= 238328a+ 3844b+ 62c+ d= 2$,

Subtracting the first from the second, 231469a+ 3461b+ 53c= 1. Since a, b, and c are integers that is a linear Diophantine equation.

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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

Gold Member
MHB
Well, to start with, since $ax^3+ bx^2+ cx+ d$ is 1 when x 19, $a(19)^3+ b(19)^2+ 19x+ d= 6859a+ 361b+ 19c+ d= 1$, And since it is 2 when x= 62, $a(62)^3+ b(62)^2+ a(62)+ d= 238328a+ 3844b+ 62c+ d= 2$,

Subtracting the first from the second, 231469a+ 3461b+ 53c= 1. Since a, b, and c are integers that is a linear Diophantine equation.
because this is a challenge question you are required to answer it fully . this is not a question for help