aoiyamada211's idea for IMO 2023 Prob 1

Exams/IMO/2023/q1.tex

@@ -56,7 +56,7 @@ \subsection*{Solution by aoiyamada211, written up by CY Fung}
 
 Case II: $d_3$ is composite.
 
-A composite number must be the product of two or more primes (not necessarily distinct). Let $d_3 = m p_b p_c $, we have $d_3 > p_b > 1 = d_1 $, $d_3 > p_c > 1 = d_1$; but there are no factors of $N$ in between $d_2 = p_a$ and $d_3$, in addition, $p_b | N$, $p_c | N$, so $p_b = d_g$ for an integer $g$ satisfying $1 < g \le 3$ and $p_c = d_h$ for an integer $h$ satisfying $1 < h \le 3$. But $p_b \ne d_3$, $p_c \ne d_3$, leads to $g = h = 2$, so $p_a = p_b = p_c$ and $d_3 = p_a^2$.
+A composite number must be the product of two or more primes (not necessarily distinct). Let $d_3 = m p_b p_c $, we have $d_3 > p_b > 1 = d_1 $, $d_3 > p_c > 1 = d_1$,  but there are no factors of $N$ in between $d_2 = p_a$ and $d_3$, in addition, $p_b | N$, $p_c | N$, so $p_b = d_g$ for an integer $g$ satisfying $1 < g \le 3$ and $p_c = d_h$ for an integer $h$ satisfying $1 < h \le 3$. But $p_b \ne d_3$, $p_c \ne d_3$, leads to $g = h = 2$, so $p_a = p_b = p_c$; for $m$, $m|N$ and $ m < d_3$, so $m=d_1 = 1$ or $m = d_2 = p_a$, the latter is impossible because if $m = p_a$, $d_3 = p_a^3$ and there would be $d_2 < p_a^2 < d_3$; thus $d_3 = p_a^2$.
 
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