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2001 IMO Shortlist Problems/N5

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Problem

Let $a > b > c > d$ be positive integers and suppose that

$ac + bd = (b + d + a - c)(b + d - a + c).$

Prove that $ab + cd$ is not prime.

Solution

Equality is equivalent to $a^2 - ac + c^2 = b^2 + bd + d^2 (1)$.

Let $ABCD$ be the quadrilateral with $AB = a$, $BC = d$, $CD = b$, $AD = c$, $\angle BAD = 60^\circ$, and $\angle BCD = 120^\circ$. Such a quadrilateral exists by $(1)$ and the Law of Cosines.

By Strong Form of Ptolemy's Theorem, we find that;

$BD^2 = \frac{(ab+cd)(ad+bc)}{ac+bd}$

and by rearrangement inequality;

$ab+cd > ac+bd > ad+bc$.

Assume $ab+cd = p$ is a prime, since $a^2 - ac + c^2 = BD^2$ is an integer $p \times \frac{ad+bc}{ac+bd}$ must be an integer but this is false since $(p,ac+bd) = 1$ and $ac+bd > ad+bc$. Thus $ab+cd$ can not be a prime.

Resources

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