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

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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)
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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>\begin{array}{rcl} nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}&=&(n-m)p^{\dfrac{1}{3}} \\
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\\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\\
  
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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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\\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\\
  
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<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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\\(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\\
  
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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>
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\\(3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})((m-n)p^{\dfrac{1}{3}})&=&(n-m)^{3}p+m^{3}r-n^{3}q\\
  
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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>
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\\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)}\\
  
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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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\\(pqr)^{\dfrac{1}{3}}&=&\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)} \end{array} </cmath>
  
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<cmath>mr^{\dfrac{1}{3}}-nq^{\dfrac{1}{3}}=(m-n)p^{\dfrac{1}{3}}</cmath>
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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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<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>
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==Solution 2==
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Assume the contrary and suppose that the cube roots of three distinct prime numbers, <math>p_1^\frac{1}{3}</math>, <math>p_2^\frac{1}{3}</math>, and <math>p_3^\frac{1}{3}</math>, form an increasing arithmetic progression. Then we can write <math>p_2^\frac{1}{3} = p_1^\frac{1}{3} + md</math> and <math>p_3^\frac{1}{3} = p_1^\frac{1}{3} + nd</math> for some positive integral <math>d, m, n</math> with <math>m < n</math>. Thus,
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<cmath>\frac{p_3^\frac{1}{3} - p_1^\frac{1}{3}}{p_2^\frac{1}{3} - p_1^\frac{1}{3}} = \frac{md}{nd} = \frac{m}{n} = r,</cmath>
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for some real <math>r</math>. Because <math>m</math> and <math>n</math> are integral, r is rational.
  
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<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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This equation rearranges to <cmath>p_3^\frac{1}{3} = rp_2^\frac{1}{3} - (r-1)p_1^\frac{1}{3}.</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>
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Note that if <math>a-b=x</math>, then <cmath>(a-b)^3 = a^3 - b^3 - 3ab(a-b) = a^3 - b^3 - 3abx</cmath> from the binomial theorem. As a result, cubing both sides of the equation gives
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<cmath>p_3 = r^3p_2 - (r-1)^3p_1 - 3r(r-1)(p_1p_2p_3)^\frac{1}{3}.    (*)</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
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''Lemma:'' The sum, product, and difference of two rational numbers are all rational, and that of a rational and irrational number are all irrational.
  
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Thus, the LHS is irrational but the RHS is rational, which causes a contradiction
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Proof: The first part is trivial if we write the fractions as <math>\frac{m}{n}</math> and <math>\frac{p}{q}</math>, and find the common denominators or simplify fractions as
 +
needed. The second part follows by contradiction: if the sum of a rational number and irrational number is rational, then the irrational number must be the difference of two rational numbers, which is rational. This is a contradiction; likewise for difference and product.
  
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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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The LHS ''(left-hand side)'' of (*) is clearly rational (in fact, integral), and so the RHS must also be rational. This is the case for the first two terms <math>r^3p_2</math> and <math>-(r-1)^3p_1</math>. But it is not the case for the third term. The number <math>(p_1p_2p_3)^\frac{1}{3}</math> is irrational because <math>p_1p_2p_3</math>, the product of three distinct primes, cannot be a perfect cube. Hence, <math>-3r(r-1)(p_1p_2p_3)^\frac{1}{3}</math>, the product of a rational and an irrational number, is irrational, and the RHS is therefore the sum of two rational terms and one irrational term. This is irrational by the lemma, and hence we have a rational equalling an irrational, contradiction. It follows that the cube roots of three distinct prime numbers cannot form an arithmetic sequence.
  
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{{USAMO box|year=1973|num-b=4|after=Last Problem}}
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== See Also ==
 +
{{USAMO box|year=1973|num-b=4|after=Last Question}}
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{{MAA Notice}}
  
 
[[Category:Olympiad Number Theory Problems]]
 
[[Category:Olympiad Number Theory Problems]]

Latest revision as of 19:18, 29 December 2019

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

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

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.

Solution 2

Assume the contrary and suppose that the cube roots of three distinct prime numbers, $p_1^\frac{1}{3}$, $p_2^\frac{1}{3}$, and $p_3^\frac{1}{3}$, form an increasing arithmetic progression. Then we can write $p_2^\frac{1}{3} = p_1^\frac{1}{3} + md$ and $p_3^\frac{1}{3} = p_1^\frac{1}{3} + nd$ for some positive integral $d, m, n$ with $m < n$. Thus, \[\frac{p_3^\frac{1}{3} - p_1^\frac{1}{3}}{p_2^\frac{1}{3} - p_1^\frac{1}{3}} = \frac{md}{nd} = \frac{m}{n} = r,\] for some real $r$. Because $m$ and $n$ are integral, r is rational.

This equation rearranges to \[p_3^\frac{1}{3} = rp_2^\frac{1}{3} - (r-1)p_1^\frac{1}{3}.\]

Note that if $a-b=x$, then \[(a-b)^3 = a^3 - b^3 - 3ab(a-b) = a^3 - b^3 - 3abx\] from the binomial theorem. As a result, cubing both sides of the equation gives \[p_3 = r^3p_2 - (r-1)^3p_1 - 3r(r-1)(p_1p_2p_3)^\frac{1}{3}. (*)\]

Lemma: The sum, product, and difference of two rational numbers are all rational, and that of a rational and irrational number are all irrational.

Proof: The first part is trivial if we write the fractions as $\frac{m}{n}$ and $\frac{p}{q}$, and find the common denominators or simplify fractions as needed. The second part follows by contradiction: if the sum of a rational number and irrational number is rational, then the irrational number must be the difference of two rational numbers, which is rational. This is a contradiction; likewise for difference and product.

The LHS (left-hand side) of (*) is clearly rational (in fact, integral), and so the RHS must also be rational. This is the case for the first two terms $r^3p_2$ and $-(r-1)^3p_1$. But it is not the case for the third term. The number $(p_1p_2p_3)^\frac{1}{3}$ is irrational because $p_1p_2p_3$, the product of three distinct primes, cannot be a perfect cube. Hence, $-3r(r-1)(p_1p_2p_3)^\frac{1}{3}$, the product of a rational and an irrational number, is irrational, and the RHS is therefore the sum of two rational terms and one irrational term. This is irrational by the lemma, and hence we have a rational equalling an irrational, contradiction. It follows that the cube roots of three distinct prime numbers cannot form an arithmetic sequence.

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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