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{ | ||
"name": "astroplate", | ||
"name": "reednel.com", | ||
"version": "1.4.1", | ||
"description": "Astro and Tailwindcss boilerplate", | ||
"author": "zeon.studio", | ||
"description": "Reed Nelson's personal website.", | ||
"author": "reednel", | ||
"license": "MIT", | ||
"packageManager": "[email protected]", | ||
"scripts": { | ||
|
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"remark-collapse": "^0.1.2", | ||
"remark-math": "^5.1.1", | ||
"remark-toc": "^8.0.1", | ||
"sharp": "^0.32.4", | ||
"swiper": "^10.1.0" | ||
"sharp": "^0.32.4" | ||
}, | ||
"devDependencies": { | ||
"@tailwindcss/forms": "^0.5.4", | ||
|
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--- | ||
title: "Fun Problems in Number Theory" | ||
meta_title: "" | ||
description: "none" | ||
date: 2023-08-07T05:00:00Z | ||
image: "" | ||
categories: ["Mathematics"] | ||
author: "Reed Nelson" | ||
tags: ["math", "puzzles"] | ||
draft: true | ||
--- | ||
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## Easy | ||
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### yeah | ||
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## Medium | ||
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### A fact | ||
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Prove that a positive integer $n$ is prime if and only if $n$ is not divisible | ||
by any prime $p$ with $1 < p \leq \sqrt{n}$. | ||
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Proving $\implies$ | ||
This follows directly from the definition of prime. | ||
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Proving $\impliedby$ | ||
Clearly if there were such a prime $p$ in range $(1, \sqrt{n}]$, n would not be prime. So then it suffices to show why the range $(\sqrt{n}, n)$ need not be checked. | ||
The largest value that can be produced by a pair in range $(1, \sqrt{n}]$ is $\sqrt{n} \cdot \sqrt{n} = n$. | ||
That means in order for a value greater than $\sqrt{n}$ to be a divisor, so too must there be a value less than $\sqrt{n}$. | ||
So if no such value less than $\sqrt{n}$ exists, no such value greater than $\sqrt{n}$ can exist. | ||
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### Facto | ||
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Prove that if $n > 4$ is composite then $n | (n - 1)!$. | ||
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An equivalent statement: if $n > 4$ is composite, then $n$ is a factor of $(n-1)!$. | ||
$n$ is a product of prime factors. | ||
If all such factors are unique in the product, then trivially they are all present in the product $(n-1)!$. | ||
If there are duplicates, they can be thought of as multiplying to unique composite numbers less than $n$, due to the fundamental theorem of arithmetic. | ||
The only time this is not true is for the composite $n = 4$, since $3!$ is divisible by no composite numbers. | ||
Hence, all prime factors of $n$ can be accounted for in $(n-1)!$ when $n > 4$, so $n$ is a factor of $(n-1)!$. | ||
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### Another Fact | ||
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Prove that if a positive integer n is a perfect square, then n | ||
cannot be written in the form $4k + 3$ for $k$ an integer. | ||
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Hint: Compute the remainder upon division by 4 of each of $(4m)^2$, $(4m + 1)^2$, $(4m + 2)^2$, and $(4m + 3)^2$. | ||
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Any positive integer of the form $4k+3$ is congruent to $3 (mod 4)$. | ||
If $n$ is even, then $n^2$ is even, and trivially cannot be of the form $4k+3$. | ||
If $n$ is odd, then $n = 2m + 1$, $m \in \mathbb{N}$. | ||
Then $(2m+1)^2 = 4m^2 + 4m + 1$, which is congruent to 3 (mod 4), so cannot be of the form $4k+3$. | ||
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### Follow-up Fact | ||
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Prove that no integer in the sequence | ||
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$11$, $111$, $1111$, $11111$, $\dots$ | ||
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is a perfect square. | ||
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Hint: $111 \dots 111 = 111 \dots 108 + 3 = 4k+ 3$. | ||
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Per the hint, all integers in the given sequence are of the form $4k+3$, and we just showed that no positive integer of that form may be a perfect square, so we are done. | ||
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### A series of fun facts | ||
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#### 3 | ||
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A number $n \in \mathbb{Z}$ is divisible by 3 if and only if the sum of the digits of $n$ is divisible by 3. | ||
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Proof: write $n = a + 10b + 100c + \dots$, where the digits of $n$ are $a, b, c, \dots$. Since $10 \equiv 1 \pmod{3}$, $n = a + 10b + 100c + \dots \equiv a, b, c, \dots \pmod{3}$. The proposition follows. | ||
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#### 5 | ||
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A number $n \in \mathbb{Z}$ is divisible by 5 if and only if the rightmost digit is 5 or 0. | ||
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Write $n = a + 10b + 100c + \dots$. | ||
Since $10 \equiv 0 \pmod{5}$, $n = a + 10b + 100c + \dots \equiv a \pmod{5}$. | ||
The only nonnegative integers less than 10 (i.e. that $a$ could be) are 5 or 0. | ||
From this, the propositon follows. | ||
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#### 9 | ||
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A number $n \in \mathbb{Z}$ is divisible by 9 if and only if the sum of the digits is. | ||
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Write $n = a + 10b + 100c + \dots$. | ||
Since $10 \equiv 1 \pmod{9}$, $n = a + 10b + 100c + \dots \equiv a + b + c + \dots \pmod{9}$. | ||
From this, the propositon follows. | ||
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#### 11 | ||
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A number $n \in \mathbb{Z}$ is divisible by 11 if and only if the difference between the sum of the digits at even indices and the ones at odd indices are congruent to $0 \pmod{11}$. | ||
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Write $n = a + 10b + 100c + \dots$. | ||
Since $10 \equiv -1 \pmod{11}$, $n = a + 10b + 100c + \dots \equiv -a + b - c + \dots \pmod{11}$. | ||
From this, the propositon follows. | ||
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Additional note: | ||
*Definition* (Palindrome): let the digits of $n \in \pmod{11}$ be of the form $abc...cba$. Then $n$ is a palindrome. | ||
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*Remark*: From the proof above, it follows that all even-length palindromes are divisible by 11. | ||
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*Conjecture*: All palindromes divisible by 11 are such that the product of their digits is a perfect square. | ||
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### One more | ||
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Prove that if $p$ is a positive integer such that both $p$ and $p^2 + 2$ are prime, then $p = 3$. | ||
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Trivially, 2 does not satisfy the conditons and 3 does. | ||
Now we consider only prime $p > 3$. | ||
Every third integer is a multiple of 3, yet no $p$ is. | ||
So for all $p$, it is either that $3 \mid p + 1$ or $3 \mid p + 2$. | ||
That is, $p \equiv 1 \pmod{3}$ or $p \equiv -1 \pmod{3}$. | ||
In either case, this gives $p^2 \equiv 1 \pmod{3}$, so $p^2 + 2 \equiv 0 \pmod{3}$. | ||
Therefore, for all primes greater than 3, $3 | p^2 + 2$, and $p^2 + 2$ is composite. | ||
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## Hard | ||
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### Fermat's Last Theorem | ||
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Prove that no triple $a, b, c, \in \mathbb{N}$ satisfy the equation $a^n + b^n = c^n$, where $n \in \mathbb{N}$ is greater than $2$. | ||
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### Collatz Conjecture | ||
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Consider the following operation on an arbitrary positive integer: | ||
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$ | ||
f(n) = \left\{ | ||
\begin{array}{lr} | ||
n / 2, & \text{if } n \equiv 0 \pmod{2}\\ | ||
3n + 1, & \text{if }n \equiv 1 \pmod{2} | ||
\end{array} | ||
\right\} | ||
$ | ||
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Prove that for all $n$, this function will eventually return 1. | ||
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### Goldbach's Conjecture | ||
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Prove that every even $n \in \mathbb{n}$ greater than 2 is the sum of two primes. |
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--- | ||
title: "Authors" | ||
title: "Drinks" | ||
meta_title: "" | ||
description: "this is meta description" | ||
--- |
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--- | ||
title: "Drink 1" | ||
meta_title: "" | ||
description: "none" | ||
date: 2022-08-30T05:00:00Z | ||
image: "" | ||
categories: ["Drink"] | ||
author: "Reed Nelson" | ||
tags: ["drink"] | ||
draft: false | ||
--- | ||
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This is a recipe. |
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--- | ||
title: "Drink 2" | ||
meta_title: "" | ||
description: "none" | ||
date: 2022-09-30T05:00:00Z | ||
image: "" | ||
categories: ["Drink2"] | ||
author: "Reed Nelson" | ||
tags: ["drink2"] | ||
draft: false | ||
--- | ||
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This is a recipe. |
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--- | ||
title: "Drink 3" | ||
meta_title: "" | ||
description: "none" | ||
date: 2022-10-30T05:00:00Z | ||
image: "" | ||
categories: ["Drink"] | ||
author: "Reed Nelson" | ||
tags: ["drink"] | ||
draft: false | ||
--- | ||
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This is a recipe. |
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--- | ||
title: "Projects" | ||
meta_title: "" | ||
description: "this is meta description" | ||
--- |
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--- | ||
title: AAG Key Exchange Implementation | ||
description: this is meta description | ||
technologies: ["Python", "SageMath"] | ||
order: 2 | ||
social: | ||
- name: github | ||
icon: FaGithub | ||
link: https://github.com/reednel/aag | ||
--- | ||
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This progect provides the first generic implementation of AAG, using the SageMath computer algebra system. This enables the direct and convenient comparison between different platform groups, which is something largely lacking in existing literature. For a more detailed explanation of AAG and this program's capabilities, see [this paper](https://github.com/reednel/aag/blob/paper/main.pdf). |
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