Difference between revisions of "1973 USAMO Problems/Problem 5"
(→Solution 2) |
(→Solution) |
||
| Line 22: | Line 22: | ||
\\(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}})(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\\ | \\(3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})((m-n)p^{\dfrac{1}{3}})&=&(n-m)^{3}p+m^{3}r-n^{3}q\\ | ||
Revision as of 22:36, 20 June 2018
Contents
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 
, 
, and 
WLOG, let 
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 integers, and 
is the common difference in the progression. Then we have
![\[\begin{array}{rcl} nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}&=&(n-m)p^{\dfrac{1}{3}} \\ \\n^{3}q-3n^{2}mq^{\dfrac{2}{3}}r^{\dfrac{1}{3}}+3nm^{2}q^{\dfrac{1}{3}}r^{\dfrac{2}{3}}-m^{3}r&=&(n-m)^{3}p\\ \\3nm^{2}q^{\dfrac{1}{3}}r^{\dfrac{2}{3}}-3n^{2}mq^{\dfrac{2}{3}}r^{\dfrac{1}{3}}&=&(n-m)^{3}p+m^{3}r-n^{3}q\\ \\(3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})(mr^{\dfrac{1}{3}}-nq^{\dfrac{1}{3}})&=&(n-m)^{3}p+m^{3}r-n^{3}q\\ \\(3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})((m-n)p^{\dfrac{1}{3}})&=&(n-m)^{3}p+m^{3}r-n^{3}q\\ \\q^{\dfrac{1}{3}}r^{\dfrac{1}{3}}p^{\dfrac{1}{3}}&=&\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)}\\ \\(pqr)^{\dfrac{1}{3}}&=&\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)} \end{array}\]](http://latex.artofproblemsolving.com/8/4/4/84471fe5badd5897cfe37e96cf3d8b202587d878.png)
Because , , 
are distinct primes, 
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, 
, 
, and 
, form an increasing arithmetic progression. Then we can write 
and 
for some positive integral 
with 
. Thus,
![\[\frac{p_3^\frac{1}{3} - p_1^\frac{1}{3}}{p_2^\frac{1}{3} - p_1^\frac{1}{3}} = \frac{md}{nd} = \frac{m}{n} = r,\]](http://latex.artofproblemsolving.com/b/4/5/b4519468c8b023d72ca044d0747929644a85e8eb.png)
for some real . Because and are integral, r is rational.
This equation rearranges to ![\[p_3^\frac{1}{3} = rp_2^\frac{1}{3} - (r-1)p_1^\frac{1}{3}.\]](http://latex.artofproblemsolving.com/f/e/3/fe3a70995bc5de4acca3aa93362800dd0d201823.png)
Note that if 
, then ![\[(a-b)^3 = a^3 - b^3 - 3ab(a-b) = a^3 - b^3 - 3abx\]](http://latex.artofproblemsolving.com/3/b/b/3bb1489870640c649e4a3a8ae89658d66bd86112.png)
from the binomial theorem. As a result, cubing both sides of the equation gives
![\[p_3 = r^3p_2 - (r-1)^3p_1 - 3r(r-1)(p_1p_2p_3)^\frac{1}{3}. (*)\]](http://latex.artofproblemsolving.com/0/d/2/0d24ff84efa87e7bce192b4ee3a34d4fabb492ed.png)
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 
and 
, 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 
and 
. But it is not the case for the third term. The number 
is irrational because 
, the product of three distinct primes, cannot be a perfect cube. Hence, 
, 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.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
