1973 USAMO Problems/Problem 5

Problem

Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression.

Let the three distinct prime number be $p$ , $q$ , and $r$

WLOG, let $p<q<r$

Assuming that the cube roots of three distinct prime numbers $can$ be three terms of an arithmetic progression.

Then,

where $m$ , $n$ are distinct integer, and d is the common difference in the progression (it's not necessary an integer)

now using the fact that $p$ , $q$ , $r$ are distinct primes, $pqr$ is not a cubic

Thus, the LHS is irrational but the RHS is rational, which causes a contradiction

Thus, the cube roots of three distinct prime numbers cannot be three terms of an arithmetic progression.

1973 USAMO (Problems)
Preceded by Problem 4	1 • 2 • 3 • 4 • 5	Followed by Last Problem
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