The following problem was posed in the March 2025 issue of Crux Mathematicorum.
OC 722 : Let $p$ and $q$ be distinct primes. Assume that the four numbers $p^{23}, p^{24}, q^{23}, q^{24}$ occur (not necessarily consecutively) in a decreasing arithmetic progression. Show that the primes $p$ and $q$ themselves also appear in that same progression.
Solution by Srikanth, Mudhitha Maths Academy, Chennai.
Show Solution
Let the common difference of the progression be $d < 0$, and list its terms as ${a_k}$ for ${k \ge 0}$. For any indices $m,n$ we have
$$ a_n – a_m = (n – m)d \quad \Rightarrow \quad d \mid (a_n – a_m). $$ In particular,
$$ d \mid (p^{24} – p^{23}), \quad d \mid (p^{23} – q^{23}), \quad d \mid (q^{24} – q^{23}). $$ First, $d$ shares no prime factor with either $p$ or $q$.
Indeed, if $p \mid d$, then $p \mid (p^{23} – q^{23})$, forcing
$p \mid q^{23}$ and hence $p = q$, a contradiction.
Thus, $\gcd(d, p) = \gcd(d, q) = 1$. Because $d \mid p^{23}(p – 1)$ and is coprime to $p^{23}$, Euclid’s lemma gives $d \mid (p – 1)$.
By symmetry we also have $d \mid (q – 1)$. Thus,
$$ p \equiv 1 \pmod{d} \quad \Rightarrow \quad p^{23} \equiv 1 \pmod{d}, $$ so
$$ d \mid (p^{23} – p). $$ The same reasoning yields $d \mid (q^{23} – q)$. Hence $p$ and $q$ both differ from their powers $p^{23}, q^{23}$ by integer multiples of $d$, and must therefore lie in the same arithmetic progression. Therefore, both $p$ and $q$ occur in the A.P.
Authors:
Srikanth is the instructor at Mudhitha Maths Academy, Chennai. He is very fond of problem solving and puzzle games. He is an assistant professor of mathematics by the week, enjoys designing software and loves teaching mathematics. Srikanth likes to play role playing or strategy video games, enjoys reading fantasy novels and collect fountain pens. His favorite subjects are geometry and category theory.