Number Theoretical Problem Proved by Group Theory. $a^{2^n}+b^{2^n}\equiv 0 \pmod{p}$ Implies $2^{n+1}|p-1$.

Group Theory Problems and Solutions in Mathematics

Problem 344

Let $a, b$ be relatively prime integers and let $p$ be a prime number.
Suppose that we have
\[a^{2^n}+b^{2^n}\equiv 0 \pmod{p}\] for some positive integer $n$.

Then prove that $2^{n+1}$ divides $p-1$.

 
LoadingAdd to solve later

Sponsored Links

Proof.

Since $a$ and $b$ are relatively prime, at least one of them is relatively prime to $p$.
Without loss of generality let us assume that $b$ and $p$ are relatively prime.

Then the given equality becomes
\begin{align*}
a^{2^n}\equiv -b^{2^n} \pmod{p} \\
\iff \left( \frac{a}{b}\right)^{2^n} \equiv -1 \pmod{p}.
\end{align*}
Taking square of both sides we obtain
\[\left( \frac{a}{b}\right)^{2^{n+1}} \equiv 1 \pmod{p}.\]


Now, we can think of these congruences as equalities of elements in the multiplicative group $(\Z/p\Z)^{\times}$ of order $p-1$:
\[ \left( \frac{a}{b}\right)^{2^n} = -1 \text{ and } \left( \frac{a}{b}\right)^{2^{n+1}} =1 \text{ in } (\Z/p\Z)^{\times}.\]

Note that the second equality yields that the order of the element $a/b$ divides $2^{n+1}$.
On the other hand, the first equality implies that any smaller power of $2$ is not the order of $a/b$.
Thus, the order of the element $a/b$ is exactly $2^{n+1}$.


In general, the order of each element divides the order of the group.
(This is a consequence of Lagrange’s theorem.)

Since the order of the group $(\Z/p\Z)^{\times}$ is $p-1$, it follows that $2^{n+1}$ divides $p-1$.
This completes the proof.


LoadingAdd to solve later

Sponsored Links

More from my site

  • Use Lagrange’s Theorem to Prove Fermat’s Little TheoremUse Lagrange’s Theorem to Prove Fermat’s Little Theorem Use Lagrange's Theorem in the multiplicative group $(\Zmod{p})^{\times}$ to prove Fermat's Little Theorem: if $p$ is a prime number then $a^p \equiv a \pmod p$ for all $a \in \Z$.   Before the proof, let us recall Lagrange's Theorem. Lagrange's Theorem If $G$ is a […]
  • Normal Subgroup Whose Order is Relatively Prime to Its IndexNormal Subgroup Whose Order is Relatively Prime to Its Index Let $G$ be a finite group and let $N$ be a normal subgroup of $G$. Suppose that the order $n$ of $N$ is relatively prime to the index $|G:N|=m$. (a) Prove that $N=\{a\in G \mid a^n=e\}$. (b) Prove that $N=\{b^m \mid b\in G\}$.   Proof. Note that as $n$ and […]
  • A One-Line Proof that there are Infinitely Many Prime NumbersA One-Line Proof that there are Infinitely Many Prime Numbers Prove that there are infinitely many prime numbers in ONE-LINE.   Background There are several proofs of the fact that there are infinitely many prime numbers. Proofs by Euclid and Euler are very popular. In this post, I would like to introduce an elegant one-line […]
  • Find the Largest Prime Number Less than One Million.Find the Largest Prime Number Less than One Million. Find the largest prime number less than one million. What is a prime number? A natural number is called a "prime number" if it is only divisible by $1$ and itself. For example, $2, 3, 5, 7$ are prime numbers, although the numbers $4,6,9$ are not. The prime numbers have always […]
  • Mathematics About the Number 2017Mathematics About the Number 2017 Happy New Year 2017!! Here is the list of mathematical facts about the number 2017 that you can brag about to your friends or family as a math geek. 2017 is a prime number Of course, I start with the fact that the number 2017 is a prime number. The previous prime year was […]
  • A Subgroup of the Smallest Prime Divisor Index of a Group is NormalA Subgroup of the Smallest Prime Divisor Index of a Group is Normal Let $G$ be a finite group of order $n$ and suppose that $p$ is the smallest prime number dividing $n$. Then prove that any subgroup of index $p$ is a normal subgroup of $G$.   Hint. Consider the action of the group $G$ on the left cosets $G/H$ by left […]
  • The Group of Rational Numbers is Not Finitely GeneratedThe Group of Rational Numbers is Not Finitely Generated (a) Prove that the additive group $\Q=(\Q, +)$ of rational numbers is not finitely generated. (b) Prove that the multiplicative group $\Q^*=(\Q\setminus\{0\}, \times)$ of nonzero rational numbers is not finitely generated.   Proof. (a) Prove that the additive […]
  • The Set of Square Elements in the Multiplicative Group $(\Zmod{p})^*$The Set of Square Elements in the Multiplicative Group $(\Zmod{p})^*$ Suppose that $p$ is a prime number greater than $3$. Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$. (a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$. (b) Determine the index $[G : S]$. (c) Assume […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Group Theory
Abelian Group problems and solutions
Abelian Normal subgroup, Quotient Group, and Automorphism Group

Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$. Let $\Aut(N)$ be the...

Close