Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.7 AIME Prove it Austria

Let $p$ and $q$ be real numbers such that the quadratic equation
$x^2 + px + q = 0$
has two real solutions $x_1$ and $x_2$ .
The following two conditions hold:
(i) The numbers $x_1$ and $x_2$ differ by 1.
(ii) The numbers $p$ and $q$ differ by 1.
Show that $p, q, x_1$ and $x_2$ are integers.

Solution

Without loss of generality, we assume $x_1 = x_2 + 1$ . By Vieta's formulas, this implies $p = -(x_1 + x_2) = -2x_2 - 1$ and $q = x_1 x_2 = x_2^2 + x_2$ .
Therefore, it is enough to check that $x_2$ has to be an integer.

**Case 1: $q = p - 1$ **
This implies $x_2^2 + x_2 = -2x_2 - 1 - 1$ and therefore $x_2^2 + 3x_2 + 2 = 0$ and $x_2 = -1$ or $x_2 = -2$ , both of which are integers.

**Case 2: $q = p + 1$ **
We find $x_2^2 + x_2 = -2x_2 - 1 + 1$ and therefore $x_2^2 + 3x_2 = 0$ and $x_2 = 0$ or $x_2 = -3$ , both of which are integers.

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