Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2001 IMO Shortlist Problems/N4"

(→‎Solution (Elementary Group Theory))
(→‎Solution (Elementary Group Theory) (And a more elementary solution given below))
 
(3 intermediate revisions by the same user not shown)
Line 20: Line 20:
 
<cmath>2^{4}\equiv16\text{ (mod }25\text{)}</cmath>
 
<cmath>2^{4}\equiv16\text{ (mod }25\text{)}</cmath>
 
<cmath>3^{4}\equiv6\text{ (mod }25\text{)}</cmath>
 
<cmath>3^{4}\equiv6\text{ (mod }25\text{)}</cmath>
−
 
−
Note: I am a different person. I want to give a easier and elementary solution. i provide it below-:
 
−
 
−
Solution-:
 
−
 
−
Claim:
 
−
 
−
<math>p^2 \nmid a^{p-1} -1</math> for any a such that <math>2 \leq a\leq p-2</math>.
 
−
 
−
Proof:
 
−
 
−
Let A= {2,3,...,p-2}.
 
−
Whenever <math>a \in A</math>, <math>p-a \in A</math>.
 
−
For the sake of contradiction, let <math>p \mid a^{p-1}-1</math> and <math>p \mid (p-a)^{p-1}-1</math>.
 
−
From the binomial theorem, we have
 
−
<math>p^2 \mid (p-a)^{p-1}-1= \dbinom{p-1}{0} \cdot p^{p-1} -...+\dbinom{p-1}{p-3} \cdot p^2 \cdot a^{p-3}-\dbinom{p-1}{p-2} \cdot p \cdot a^{p-2}+\dbinom{p-1}{p-1} \cdot a^{p-1}-1</math>.
 
−
As <math>p^2 \mid  \dbinom{p-1}{0} \cdot p^{p-1} -...+\dbinom{p-1}{p-3} \cdot p^2 \cdot a^{p-3}</math> and <math>p^2 \mid a^{p-1}-1= \dbinom{p-1}{p-1} \cdot a^{p-1}-1</math>, <math>p^2 \mid -\dbinom{p-1}{p-2} \cdot p \cdot a^{p-2}</math> so that <math>p^2 \mid \dbinom{p-1}{p-2} \cdot p \cdot a^{p-2} = \dbinom{p-1}{1} \cdot p \cdot a^{p-2}= (p-1) \cdot p \cdot a^{p-2}</math> which further implies that <math>p \mid (p-1) \cdot a^{p-2}</math>, which is a clear contradiction as gcd(p,p-1)= 1 and gcd(p,a)=1, so that gcd(p,<math>a^{p-2}</math>)= 1.
 
−
 
−
Thus, our claim is proved.
 
−
 
−
Note that there is at least one pair of consecutive integers in the set {2,3,...,p-2} for any prime <math>p \geq 5</math>, thus ending our proof.
 
  
 
== Resources ==
 
== Resources ==

Latest revision as of 21:58, 7 June 2025

Problem

Let $p \geq 5$ be a prime number. Prove that there exists an integer $a$ with $1 \leq a \leq p - 2$ such that neither $a^{p - 1} - 1$ nor $(a + 1)^{p - 1} - 1$ is divisible by $p^2$.

Solution (Elementary Group Theory)

Let us work in the group $(\mathbb{Z}/p^2\mathbb{Z})^{\times}$.

Note that the order of this group is $\phi({p^2})=p^2-p$. Since $(\mathbb{Z}/p^2\mathbb{Z})^{\times}$ is cyclic, we know that it is isomorphic to $(\mathbb{Z}/(p^2-p)\mathbb{Z})^{+}$.

We can then conclude how many residues $n$ there are such that $(p-1)n\equiv0\text{ (mod }p^2-p\text{)}$. For some arbitrary natural number $k$, one has: \[(p-1)n=k(p^2-p)\] \[n=kp\] Therefore there are only $p$ possible values of $n$. It's also notable that if this is true for $n_0$, then it is also true for $p^2-p-n_0$ and $2n_0$. This implies that there are also $p$ different values of a such that $a^{p-1}\equiv 1\text{ (mod }p^2\text{)}$. It also implies that if and only if $a_0$ has this property, so does $\frac{1}{a_0}$, $a_0^2$ and $\frac{1}{a_0^2}$ when working in $(\mathbb{Z}/p^2\mathbb{Z})^{\times}$. The squares also have their inverses which are guaranteed to have the property.

There exists no numbers $1<m_1,m_2\leq p-2$ such that $m_1m_2\equiv1\text{ (mod }p^2\text{)}$ as $(p-2)(p-2)<p^2$ and $2^2>1$. Therefore, at most half of the values where $a_0^{p-1}\equiv 1\text{ (mod }p^2\text{)}$ are in range $1<a_0\leq p-2$. We can again remove some of the potential values by squaring all $a_0\geq\sqrt{p}$ and taking their inverse. As long as no more than half of the values in that range can have the required property, there must be a pair $a$ and $a+1$ that satisfy the requirements by the pigeonhole principle. \[\frac{p-\sqrt(p-2)-1}{2}\leq\frac{p-3}{2}\] \[2\leq(\sqrt(p-2))\] \[p\geq6\]

Lastly, you must show that there is a pair for $p=5$. This is satisfied by $a=2$. \[2^{4}\equiv16\text{ (mod }25\text{)}\] \[3^{4}\equiv6\text{ (mod }25\text{)}\]

Resources

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_IMO_Shortlist_Problems/N4&oldid=249974"