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> | ||
| + | |||
| + | WLOG, let <math>p<q<r</math> | ||
| + | |||
| + | Assuming that the cube roots of three distinct prime numbers <math>can</math> be three terms of an arithmetic progression. | ||
| + | |||
| + | Then, | ||
| + | |||
| + | <cmath>q^{\dfrac{1}{3}}=p^{\dfrac{1}{3}}+md</cmath> | ||
| + | |||
| + | <cmath>r^{\dfrac{1}{3}}=p^{\dfrac{1}{3}}+nd</cmath> | ||
| + | |||
| + | 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) | ||
| + | |||
| + | <cmath>nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}=(n-m)p^{\dfrac{1}{3}}-----(1)</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>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>(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>nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}=(n-m)p^{\dfrac{1}{3}}-----(1)</cmath> | ||
| + | |||
| + | <cmath>mr^{\dfrac{1}{3}}-nq^{\dfrac{1}{3}}=(m-n)p^{\dfrac{1}{3}}</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-----(1) and (2)</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>(pqr)^{\dfrac{1}{3}}=\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q)}{(3mn)(m-n)}</cmath> | ||
| + | |||
| + | now using the fact that <math>p</math>, <math>q</math>, <math>r</math> are distinct primes, <math>pqr</math> 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. | ||
{{USAMO box|year=1973|num-b=4|after=Last Problem}} | {{USAMO box|year=1973|num-b=4|after=Last Problem}} | ||
[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] | ||
Revision as of 18:39, 10 August 2009
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 
, 
, and 
WLOG, let 
Assuming that the cube roots of three distinct prime numbers 
be three terms of an arithmetic progression.
Then,
![\[q^{\dfrac{1}{3}}=p^{\dfrac{1}{3}}+md\]](http://latex.artofproblemsolving.com/4/a/0/4a071d3477ec5ceddeea667cf58e0f20682d28f9.png)
![\[r^{\dfrac{1}{3}}=p^{\dfrac{1}{3}}+nd\]](http://latex.artofproblemsolving.com/c/f/8/cf8594129e79769cafaee3b4da58837178f92b46.png)
where 
, 
are distinct integer, and d is the common difference in the progression (it's not necessary an integer)
![\[nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}=(n-m)p^{\dfrac{1}{3}}-----(1)\]](http://latex.artofproblemsolving.com/0/d/1/0d1ac1d24f317a3ac37bffb58d996228e72eb9fe.png)
![\[n^{3}q-3n^{2}mq^{\dfrac{2}{3}}r^{\dfrac{1}{3}}+3nm^{2}q^{\dfrac{1}{3}}r^{\dfrac{2}{3}}-m^{3}r=(n-m)^{3}p\]](http://latex.artofproblemsolving.com/c/e/e/cee1c5a968f5a721f49041a0ec041835de390d9c.png)
![\[3nm^{2}q^{\dfrac{1}{3}}r^{\dfrac{2}{3}}-3n^{2}mq^{\dfrac{2}{3}}r^{\dfrac{1}{3}}=(n-m)^{3}p+m^{3}r-n^{3}q\]](http://latex.artofproblemsolving.com/8/3/a/83afed55a83c51ef649ad7485847b8a9e5a68878.png)
![\[(3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})(mr^{\dfrac{1}{3}}-nq^{\dfrac{1}{3}})=(n-m)^{3}p+m^{3}r-n^{3}q-----(2)\]](http://latex.artofproblemsolving.com/6/9/c/69cce14ced609162fdf598a6abb8802c36cc3b92.png)
![\[mr^{\dfrac{1}{3}}-nq^{\dfrac{1}{3}}=(m-n)p^{\dfrac{1}{3}}\]](http://latex.artofproblemsolving.com/4/2/8/428f01cecb247a355e3b582e1330513eec27a6ef.png)
![\[(3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})((m-n)p^{\dfrac{1}{3}})=(n-m)^{3}p+m^{3}r-n^{3}q-----(1) and (2)\]](http://latex.artofproblemsolving.com/0/a/b/0ab4f3fab33deec89fdc0723acde439d62241cb5.png)
![\[q^{\dfrac{1}{3}}r^{\dfrac{1}{3}}p^{\dfrac{1}{3}}=\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q)}{(3mn)(m-n)}\]](http://latex.artofproblemsolving.com/2/e/1/2e1a166e2dbe101939657860bd211497533c6fe4.png)
![\[(pqr)^{\dfrac{1}{3}}=\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q)}{(3mn)(m-n)}\]](http://latex.artofproblemsolving.com/a/0/9/a09250d5ba3b988db768b184e4d220fc6586d12e.png)
now using the fact that , , are distinct primes, 
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 (Problems • Resources) | ||
| Preceded by Problem 4 |
Followed by Last Problem | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
