Competition problem — Number theory

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Number theory Difficulty 8.7 Shortlist Prove it Saudi Arabia

By $\text{rad}(x)$ we denote the product of all distinct prime factors of a positive integer $n$ . Given $a \in \mathbb{N}$ , a sequence $(a_n)$ is defined by $a_0 = a$ and $a_{n+1} = a_n + \text{rad}(a_n)$ for all $n \ge 0$ . Prove that there exists an index $n$ for which $\frac{a_n}{\text{rad}(a_n)} = 2022$ .

Solution

Solution. To prove that $MP + NQ = BR$ , by adding $RC$ to both sides it is equivalent to prove $CQ + MP = BC$ . We start by defining point $T$ on the segment $CQ$ such that $BC = CT$ , then it is sufficient to prove that $QT = PM$ , since triangles $BTC$ and $NRC$ are isosceles. One can get that $BT$ is parallel to $NR$ so
$\angle BTC = \angle RNC = \angle NRC = \angle BPC,$
thus $BTPC$ is cyclic. Also we have that
$\angle CNA = 180^\circ - \angle NRC = 180^\circ - \angle BPC = \angle CPA,$
this implies that $PNCA$ is cyclic.

$\angle TPQ = \angle TCB = \angle BQC = \angle TQP,$
thus $TQ = TP$ . Now by angle chasing,
$\angle CPA = 180^\circ - \angle BPC = 180^\circ - \angle BTC = 180^\circ - \angle TBC = \angle TPC,$
thus we have $\angle CPQ = \angle TPC$ and $\angle PCT = \angle PAM$ , which implies that $TPC$ and $MPA$ are congruent, since $PC = AP$ . In conclusion, $TP = MP = TQ$ , which finishes the proof. □

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