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,

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

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

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

$nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}=(n-m)p^{\dfrac{1}{3}}-----(1)$

$n^{3}q-3n^{2}mq^{\dfrac{2}{3}}r^{\dfrac{1}{3}}+3nm^{2}q^{\dfrac{1}{3}}r^{\dfrac{2}{3}}-m^{3}r=(n-m)^{3}p$

$3nm^{2}q^{\dfrac{1}{3}}r^{\dfrac{2}{3}}-3n^{2}mq^{\dfrac{2}{3}}r^{\dfrac{1}{3}}=(n-m)^{3}p+m^{3}r-n^{3}q$

$(3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})(mr^{\dfrac{1}{3}}-nq^{\dfrac{1}{3}})=(n-m)^{3}p+m^{3}r-n^{3}q-----(2)$

$nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}=(n-m)p^{\dfrac{1}{3}}-----(1)$

$mr^{\dfrac{1}{3}}-nq^{\dfrac{1}{3}}=(m-n)p^{\dfrac{1}{3}}$

$(3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})((m-n)p^{\dfrac{1}{3}})=(n-m)^{3}p+m^{3}r-n^{3}q-----(1) and (2)$

$q^{\dfrac{1}{3}}r^{\dfrac{1}{3}}p^{\dfrac{1}{3}}=\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)}$

$(pqr)^{\dfrac{1}{3}}=\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)}$

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.

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)
Preceded by Problem 4	Followed by Last Question
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