Difference between revisions of "1973 USAMO Problems/Problem 5"
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Thus, the cube roots of three distinct prime numbers cannot be three terms of an arithmetic progression. | 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 | + | == See Also == |
| + | {{USAMO box|year=1973|num-b=4|after=Last Question}} | ||
[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] | ||
Revision as of 07:28, 16 September 2012
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/3/2/9/3297eac3bc9728d3ba50944f469f1659ff8607e4.png)
![\[(pqr)^{\dfrac{1}{3}}=\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)}\]](http://latex.artofproblemsolving.com/3/e/5/3e5e48448312226d55cf03a7350dc5ae8177e2fb.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.
See Also
| 1973 USAMO (Problems • Resources) | ||
| Preceded by Problem 4 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
