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Difference between revisions of "1989 IMO Problems/Problem 5"

(→‎Solution)
m (→‎Solution 2)
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==Solution 2==
 
==Solution 2==
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By Chinese Remainder theorem, there exists <math>x</math> such that:
+
By the [[Chinese Remainder Theorem]], there exists <math>x</math> such that:
 
<math>x \equiv -1\;mod\;p_1 q_1\\  
 
<math>x \equiv -1\;mod\;p_1 q_1\\  
 
x \equiv -2\;mod\;p_2 q_2\\
 
x \equiv -2\;mod\;p_2 q_2\\
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...\newline
 
...\newline
 
x \equiv -n\;mod\;p_n q_n</math>
 
x \equiv -n\;mod\;p_n q_n</math>
−
Where <math>p_1, p_2, ..., p_n, q_1, q_2, ..., q_n</math> are distinct primes.  
+
where <math>p_1, p_2, ..., p_n, q_1, q_2, ..., q_n</math> are distinct primes.  
−
The n consecutive numbers <math>x+1, x+2, ..., x+n</math> each have at least two prime factors, so none of them can be expressed as an integral power of a prime.  
+
The <math>n</math> consecutive numbers <math>x+1, x+2, ..., x+n</math> each have at least two prime factors, so none of them can be expressed as an integral power of a prime.  
  
 
== See Also == {{IMO box|year=1989|num-b=4|num-a=6}}
 
== See Also == {{IMO box|year=1989|num-b=4|num-a=6}}
  
 
[[Category:Olympiad Number Theory Problems]]
 
[[Category:Olympiad Number Theory Problems]]

Revision as of 22:30, 29 July 2024

Problem

Prove that for each positive integer $n$ there exist $n$ consecutive positive integers none of which is an integral power of a prime number.

Solution 1

There are at most $1+\sqrt[2]{n}+\sqrt[3]{n}+\sqrt[4]{n}+...+\sqrt[\left\lfloor \log_2(n)\right\rfloor]{n} \leq 1+ \sqrt n log_2(n)$ 'true' powers $m^k , k\geq 2$ in the set $\{1,2,...,n\}$. So when $p(n)$ gives the amount of 'true' powers $\leq n$ we get that $\lim_{n \to \infty} \frac{p(n)}{n} = 0$.

Since also $\lim_{n \to \infty} \frac{\pi(n)}{n} = 0$, we get that $\lim_{n \to \infty} \frac{p(n)+\pi(n)}{n} = 0$. Now assume that there is no 'gap' of lenght at least $k$ into the set of 'true' powers and the primes. Then this would give that $\frac{p(n)+\pi(n)}{n} \geq \frac{1}{k}$ for all $n$ in contrary to the above (at east this proves a bit more).

Edit: to elementarize the $\lim_{n \to \infty} \frac{\pi(n)}{n} = 0$ part: Look $\mod (k+1)!$. Then all numbers in the residue classes $2,3,4,...,k+1$ are not primes (except the smallest representants sometimes). So when there wouldn't exist a gap of length $k$, there has to be a 'true' power in each of these gaps of the prime numbers, so at least one power each $(k+1)!$ numbers, again contradicting $\lim_{n \to \infty} \frac{p(n)}{n} = 0$.

This solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [1]

Solution 2

By the Chinese Remainder Theorem, there exists $x$ such that: $x \equiv -1\;mod\;p_1 q_1\\ x \equiv -2\;mod\;p_2 q_2\\ x \equiv -3\;mod\;p_3 q_3\\ ...\newline x \equiv -n\;mod\;p_n q_n$ where $p_1, p_2, ..., p_n, q_1, q_2, ..., q_n$ are distinct primes. The $n$ consecutive numbers $x+1, x+2, ..., x+n$ each have at least two prime factors, so none of them can be expressed as an integral power of a prime.

See Also

1989 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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