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All-Soviet-Union Mathematical Olympiad

Algebra Difficulty 4.7 AIME Prove it Soviet Union

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

Solution

Solution:
If there were such values, then subtract the equation with $x = 19$ from the equation with $x = 62$ to get:
$a(62^3 - 19^3) + b(62^2 - 19^2) + c(62 - 19) = 1.$
But the left hand side is divisible by $62 - 19 = 43$ , contradiction.

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