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.

Solution

Assume that the cube roots of three distinct prime numbers can be three terms of an arithmetic progression. Let the three distinct prime numbers be $p$ , $q$ , and $r.$ WLOG, let $p<q<r.$

Then,

$q^{\dfrac{1}{3}}=p^{\dfrac{1}{3}}+md$

$r^{\dfrac{1}{3}}=p^{\dfrac{1}{3}}+nd$

where $m$ , $n$ are distinct integers, and $d$ is the common difference in the progression. Then we have

$\begin{array}{rcl} nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}&=&(n-m)p^{\dfrac{1}{3}} \\ \\n^{3}q-3n^{2}mq^{\dfrac{2}{3}}r^{\dfrac{1}{3}}+3nm^{2}q^{\dfrac{1}{3}}r^{\dfrac{2}{3}}-m^{3}r&=&(n-m)^{3}p\\ \\3nm^{2}q^{\dfrac{1}{3}}r^{\dfrac{2}{3}}-3n^{2}mq^{\dfrac{2}{3}}r^{\dfrac{1}{3}}&=&(n-m)^{3}p+m^{3}r-n^{3}q\\ \\(3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})(mr^{\dfrac{1}{3}}-nq^{\dfrac{1}{3}})&=&(n-m)^{3}p+m^{3}r-n^{3}q\\ \\nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}&=&(n-m)p^{\dfrac{1}{3}}\\ \\mr^{\dfrac{1}{3}}-nq^{\dfrac{1}{3}}&=&(m-n)p^{\dfrac{1}{3}}\\ \\(3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})((m-n)p^{\dfrac{1}{3}})&=&(n-m)^{3}p+m^{3}r-n^{3}q\\ \\q^{\dfrac{1}{3}}r^{\dfrac{1}{3}}p^{\dfrac{1}{3}}&=&\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)}\\ \\(pqr)^{\dfrac{1}{3}}&=&\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)} \end{array}$

Because $p$ , $q$ , $r$ are distinct primes, $pqr$ is not a perfect cube. Thus, the LHS is irrational but the RHS is rational, which is a contradiction. So, the cube roots of three distinct prime numbers cannot be three terms of an arithmetic progression.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1973 USAMO (Problems • Resources)
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1973 USAMO Problems/Problem 5

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