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Difference between revisions of "1973 USAMO Problems/Problem 5"

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(→‎Solution)
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==Solution==
 
==Solution==
  
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Let the three distinct prime number be <math>p</math>, <math>q</math>, and <math>r</math>
 
  
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WLOG, let <math>p<q<r</math>
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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 <math>p</math>, <math>q</math>, and <math>r.</math> WLOG, let <math>p<q<r.</math>
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Assuming that the cube roots of three distinct prime numbers <math>can</math> be three terms of an arithmetic progression.
 
  
 
Then,  
 
Then,  
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<cmath>r^{\dfrac{1}{3}}=p^{\dfrac{1}{3}}+nd</cmath>
 
<cmath>r^{\dfrac{1}{3}}=p^{\dfrac{1}{3}}+nd</cmath>
  
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where <math>m</math>, <math>n</math> are distinct integer, and d is the common difference in the progression (it's not necessary an integer)
+
where <math>m</math>, <math>n</math> are distinct integers, and <math>d</math> is the common difference in the progression. Then we have
  
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<cmath>nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}=(n-m)p^{\dfrac{1}{3}}-----(1)</cmath>
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<cmath>nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}=(n-m)p^{\dfrac{1}{3}}</cmath>
  
 
<cmath>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</cmath>
 
<cmath>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</cmath>
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<cmath>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</cmath>
 
<cmath>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</cmath>
  
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<cmath>(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)</cmath>
+
<cmath>(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</cmath>
  
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<cmath>nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}=(n-m)p^{\dfrac{1}{3}}-----(1)</cmath>
+
<cmath>nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}=(n-m)p^{\dfrac{1}{3}}</cmath>
  
 
<cmath>mr^{\dfrac{1}{3}}-nq^{\dfrac{1}{3}}=(m-n)p^{\dfrac{1}{3}}</cmath>
 
<cmath>mr^{\dfrac{1}{3}}-nq^{\dfrac{1}{3}}=(m-n)p^{\dfrac{1}{3}}</cmath>
  
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<cmath>(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)</cmath>
+
<cmath>(3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})((m-n)p^{\dfrac{1}{3}})=(n-m)^{3}p+m^{3}r-n^{3}q</cmath>
  
 
<cmath>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)}</cmath>
 
<cmath>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)}</cmath>
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<cmath>(pqr)^{\dfrac{1}{3}}=\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)}</cmath>
 
<cmath>(pqr)^{\dfrac{1}{3}}=\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)}</cmath>
  
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now using the fact that <math>p</math>, <math>q</math>, <math>r</math> are distinct primes, <math>pqr</math> is not a cubic
+
Because <math>p</math>, <math>q</math>, <math>r</math> are distinct primes, <math>pqr</math> 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.
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Thus, the LHS is irrational but the RHS is rational, which causes a contradiction
 
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Thus, the cube roots of three distinct prime numbers cannot be three terms of an arithmetic progression.
 
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{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 20:19, 20 June 2013

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

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

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

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

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

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

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

\[(3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})((m-n)p^{\dfrac{1}{3}})=(n-m)^{3}p+m^{3}r-n^{3}q\]

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

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

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.

See Also

1973 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Last Question
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions
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