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

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<cmath>3^{4}\equiv6\text{ (mod }25\text{)}</cmath>
 
<cmath>3^{4}\equiv6\text{ (mod }25\text{)}</cmath>
  
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Note: I am a different person. I want to give an easier and elementary solution. i provide it below-:
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Note: I am a different person. I want to provide a more elementary solution. I provide it below-:
  
 
Solution-:
 
Solution-:

Revision as of 07:09, 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) (And a more elementary solution given below)

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 $a=2$. \[2^{4}\equiv16\text{ (mod }25\text{)}\] \[3^{4}\equiv6\text{ (mod }25\text{)}\]

Note: I am a different person. I want to provide a more elementary solution. I provide it below-:

Solution-:

Claim:

$p^2 \nmid a^{p-1} -1$ for any a such that $2 \leq a\leq p-2$.

Proof:

Let A= {2,3,...,p-2}. Whenever $a \in A$, $p-a \in A$. For the sake of contradiction, let $p \mid a^{p-1}-1$ and $p \mid (p-a)^{p-1}-1$. From the binomial theorem, we have $p^2 \mid (p-a)^{p-1}-1= \dbinom{p-1}{0} \cdot p^{p-1} -...+\dbinom{p-1}{p-3} \cdot p^2 \cdot a^{p-3}-\dbinom{p-1}{p-2} \cdot p \cdot a^{p-2}+\dbinom{p-1}{p-1} \cdot a^{p-1}-1$. As $p^2 \mid \dbinom{p-1}{0} \cdot p^{p-1} -...+\dbinom{p-1}{p-3} \cdot p^2 \cdot a^{p-3}$ and $p^2 \mid a^{p-1}-1= \dbinom{p-1}{p-1} \cdot a^{p-1}-1$, $p^2 \mid -\dbinom{p-1}{p-2} \cdot p \cdot a^{p-2}$ so that $p^2 \mid \dbinom{p-1}{p-2} \cdot p \cdot a^{p-2} = \dbinom{p-1}{1} \cdot p \cdot a^{p-2}= (p-1) \cdot p \cdot a^{p-2}$ which further implies that $p \mid (p-1) \cdot a^{p-2}$, which is a clear contradiction as gcd(p,p-1)= 1 and gcd(p,a)=1, so that gcd(p,$a^{p-2}$)= 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 $p \geq 5$, thus ending our proof.

Resources

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