Articles of number theory

How to find all binary numbers in base $10$ s.t. that its divisible by its own numerical value in base $2$?

Consider $N=1010$. Its numerical value in base-$2$ is $2^3+2=10$ and $10 \mid 1010$. How to find all positive integers like $N$? Its easy to see that $n$ is in the form $10^{n_1}+10^{n_2}+…+10^{n_k}$, where $n_1>n_2>…>n_k \ge 0$. We need $$ 2^{n_1}+2^{n_2}+…+2^{n_k} \mid 10^{n_1}+10^{n_2}+…+10^{n_k} $$ I’ll be very grateful if you could share your thoughts on this […]

Applying Extended Euclidean Algorithm for Galois Field to Find Multiplicative Inverse

I was trying to apply the Extended Euclidean Algorithm for Galois Field. Among the many resources available, I found the methodology outlined in this document easy to grasp. The above works fine when applied to numbers. Now, for $\textit{GF}(2^3)$, if I take the polynomial $x^2$ and the irreducible polynomial $P(x) = x^3 + x + […]

Solving quadratic equations in modular arithmetic

Is there a general way to solve quadratic equations modulo n? I know how to use Legendre and Jacobi symbols to tell me if there’s a solution, but I don’t know how to get a solution without resorting to guesswork. Some examples of my problems: $x^2 = 8$ mod 2009 $x^2 + 3x + 1 […]

Why these two problems lead to same answers?

Suppose these two problems: Problem 1: $$\min_{X,P} \quad\max_{1\leq l\leq L-1} \quad {|\sum_{1\leq i\leq N_p}^{N_p}x_ie^{\frac{2\pi l}{N}p_i}| \over {\sum_{i=1}^{N_p} x_i^2}} \quad \equiv \quad \min_{X,P} \quad\max_{1\leq l\leq L-1} \quad {\sum_{1\leq i,j\leq N_p}^{N_p}x_ix_je^{\frac{2\pi l}{N}(p_i-p_j)} \over {(\sum_{i=1}^{N_p} x_i^2})^2} $$ Problem 2: $$\min_{P} \quad\max_{1\leq l\leq L-1} \quad |\sum_{1\leq i\leq N_p}^{N_p}e^{\frac{2\pi l}{N}p_i}| \quad \equiv \quad \min_{P} \quad\max_{1\leq l\leq L-1} \quad \sum_{1\leq i,j\leq […]

Show that $\left| \sqrt2-\frac{h}{k} \right| \geq \frac{1}{4k^2},$ for any $k \in \mathbb{N}$ and $h \in \mathbb{Z}$.

Show that $$\left| \sqrt2-\frac{h}{k} \right| \geq \frac{1}{4k^2},$$ for any $k \in \mathbb{N}$ and $h \in \mathbb{Z}$. I tried many different ways to expand left side and estimate it but always got stuck at some point.

How prove this $|\{n\sqrt{3}\}-\{n\sqrt{2}\}|>\frac{1}{20n^3}$

Prove that $$|\{n\sqrt{3}\}-\{n\sqrt{2}\}|>\dfrac{1}{20n^3}$$ let $t=\{n\sqrt{2}\}-\{n\sqrt{3}\}$ and $k=[n\sqrt{3}]-[n\sqrt{2}]$ then we have $$t=k-(\sqrt{3}-\sqrt{2})n=k-\sqrt{5-2\sqrt{6}}n\neq 0$$ so \begin{align*} t&=\dfrac{(k-(\sqrt{3}-\sqrt{2})n)(k-(\sqrt{3}+\sqrt{2})n)(k+(\sqrt{3}-\sqrt{2})n)(k+(\sqrt{3}+\sqrt{2})n)}{(k-(\sqrt{3}+\sqrt{2})n)(k+(\sqrt{3}-\sqrt{2})n)(k+(\sqrt{3}+\sqrt{2})n)}\\ &=\dfrac{k^4-10k^2n^2+n^4}{(t-2\sqrt{2}n)(t+2(\sqrt{3}-\sqrt{2})n)(t+2\sqrt{3}n)} \end{align*} notice that $$|t-2\sqrt{2}n|\le 2\sqrt{2}n+\dfrac{1}{20}\le(2\sqrt{2}+\dfrac{1}{20})n$$ $$|t+2(\sqrt{3}-\sqrt{2})n|\le2(\sqrt{3}-\sqrt{2})n+\dfrac{1}{20}\le(2\sqrt{3}-2\sqrt{2}+\dfrac{1}{20})n$$ $$|t+2\sqrt{3}n|\le2\sqrt{3}n+\dfrac{1}{20}\le(2\sqrt{3}+\dfrac{1}{20})n$$ so $$|t|\ge\dfrac{1}{(2\sqrt{2}+\dfrac{1}{20})(2\sqrt{3}+2\sqrt{2}-\dfrac{1}{20})(2\sqrt{3}+\dfrac{1}{20})n^3}>\dfrac{1}{7n^3}$$ so $$|t|\ge\min{\left(\dfrac{1}{20},\dfrac{1}{7n^3}\right)}\ge\dfrac{1}{20n^3}$$ \ This post https://math.stackexchange.com/questions/465419/how-prove-this-t-2-sqrt2n-le2-sqrt2n-frac120 is not true? so This methods is wrong, so How prove it? Thank you

Positive integer solutions to $\frac{(x+y)^{x+y}(y+z)^{y+z}(x+z)^{x+z}}{x^{2x}y^{2y}z^{2z}}=2016^m$

Do there exist positive integers $x,y,z$ and positive rational number $m$ such that: $$\frac{(x+y)^{x+y}(y+z)^{y+z}(x+z)^{x+z}}{x^{2x}y^{2y}z^{2z}}=2016^m$$

Let $p > 3$ be a prime number. Show that $x^2 \equiv −3\mod p$ is solvable iff $p\equiv 1\mod 6$.

Let $p > 3$ be a prime number. Show that $x^2 \equiv −3\mod p$ is solvable iff $p\equiv 1\mod 6$. My try is let $a$ be a solution of $x^2 \equiv -3 \mod p$. so $a^{p-1} \equiv 1\mod p$. This implies $$(-1)^{\frac{p-1}2}≡(a^2)^\frac{p-1}2 \mod p.$$ since $p$ is odd, $\frac{p-1}2$ must be even, so $4|(p-1)$. conversely, […]

Solving the congruence $3x^2 + 6x + 1 \equiv 0 \pmod {19}$

This question already has an answer here: solve $3x^2 + 6x +1 \equiv 0 \pmod {19}$ 2 answers

Pythagorean triples with additional parameters

I want to find solution in $\mathbb{Z}$ to the following quadratic Diophantene equation: $$na^2 + kb^2 = c^2$$ where $n,k,a,b,c \in \mathbb{Z}$, $n,k > 0$ and $(n,k) = 1$ I know that for some this won’t work for all values of $n$ and $k$ that satisfy the upper condition, but anyone know what will be […]