Articles of elementary number theory

Is every Mersenne prime of the form : $x^2+3 \cdot y^2$?

How to prove or disprove following statement : Conjecture : Every Mersenne prime number can be uniquely written in the form : $x^2+3 \cdot y^2$ , where $\gcd(x,y)=1$ and $x,y \geq 0$ Since $M_p$ is an odd number it follows that : $M_p \equiv 1 \pmod 2$ According to Fermat little theorem we can write […]

Describe all integers a for which the following system of congruences (with one unknown x) has integer solutions:

$$x\equiv a \pmod {100}$$ $$x\equiv a^2 \pmod {35}$$ $$x\equiv 3a-2 \pmod {49}$$ I’m trying to solve this system of congruences, but I’m only familiar with a method for solving when the mods are pairwise coprime. I feel like there ought to be some simplification I can make to reduce it to a system where they […]

Given a fixed prime number $p$ and fixed positive integers $a$ and $k$; Find all positive integers $n$ such that $p^k \mid a^n-1$

The question is the natural generalization of the Find all positive integers solutions such that $3^k$ divides $2^n-1$ . Let $p$ to be a prime number and let $k$ & $a$ be fixed natural numbers. Find all natural numbers $n$, such that $p^k \mid a^n-1$?

If $\gcd(a,b)=1$, is $\gcd(a^x-b^x,a^y-b^y)=a^{\gcd(x,y)}-b^{\gcd(x,y)}$?

If $\gcd(a,b)=1$, is it true that $$\gcd(a^x-b^x,a^y-b^y)=a^{\gcd(x,y)}-b^{\gcd(x,y)}\;?$$ I know that $a^{\gcd(x,y)}-b^{\gcd(x,y)}\mid a^x-b^x$ and $a^{\gcd(x,y)}-b^{\gcd(x,y)}|a^y-b^y$, so I thought of something like let $n$ be a divisor of $a^x-b^x$ and $a^y-b^y$, then $n$ must also be a divisor of $a^{\gcd(x,y)}-b^{\gcd(x,y)}$ but I am stuck.

Number of decimal places to be considered in division

This must be a basic question. But i need some help. What is the number of decimal places that needs to be considered normally in division operations in order to represent the dividend value as a multiple of divisor and quotient by rounding off. Example: Say i want to divide 3475934 and 3475935 by 65536. […]

Find all the numbers $n$ such that $\frac{12n-6}{10n-3}$ can't be reduced.

Find all the numbers $n$ such that $\frac{12n-6}{10n-3}$ can’t be reduced. Attempt: It can’t be reduced when $\gcd(12n-6,10n-3)=1$ Here $(a,b)$ denotes $\gcd(a,b)$ $$(12n-6,10n-3)=(12n-6,2n-3)=(12n-6,12n-18)=(12n-6,12)$$ $\Longrightarrow$ It can’t be redused when $12\nmid 12n-6$ i.e when $12n\not\equiv 6\pmod{12}$ Theorem: for $ax\equiv b \pmod n$ there is a solution iff $d\mid b$ where $d=\gcd(a,n)$. In this case $\gcd(12,12)=12$ so […]

Bijection between natural numbers $\mathbb{N}$ and natural plane $\mathbb{N} \times \mathbb{N}$

This question already has an answer here: bijection between $\mathbb{N}$ and $\mathbb{N}\times\mathbb{N}$ [duplicate] 3 answers

Numbers of the form $\frac{xyz}{x+y+z}$

It is easy to show that every natural number $n$ can be written as $n = \frac{xy}{x+y}$ with $x,y\in \mathbb{N}$ by setting $x = y = 2n$. Now I experimented a little bit with numbers of the form $n=\frac{xyz}{x+y+z}$ and it seems that every natural number $n$ can be written this way with $x,y,z \in […]

Last 3 digits of $7^{12341}$

I know that I need to reduce $7^{12341} \pmod {1000}$ By Euler I have $7^{\phi(1000)}\equiv 7^{400}\equiv1\pmod{1000}$ That leaves me with the monster $7^{341}\pmod{1000}$ Is there a way to reduce this smoothly without working $7^2, 7^4, 7^8$ etc manually ?

Chinese Remainder Theorem Problem

I’m working on some Chinese Remainder problems and it doesn’t seem like I have the procedure down correctly. I’ll list the steps I’m taking so hopefully someone can spot what it is I’m doing wrong. Find the least nonnegative solution of each system of congruences below. $x \equiv 3 \space mod \space 4$ $x \equiv […]