1973 USAMO Problems/Problem 5

Revision as of 01:51, 27 February 2012 by Kubluck (talk | contribs) (Solution)

Problem

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

Solution

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.

1973 USAMO (ProblemsResources)
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Problem 4
Followed by
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