update ent

src/content/blog/ent.md

@@ -12,15 +12,20 @@ draft: true
 
 ## Easy
 
+See https://mathriddles.williams.edu/?cat=694
+
 ### yeah
 
 ## Medium
 
-### A fact
+### 2.1
 
 Prove that a positive integer $n$ is prime if and only if $n$ is not divisible
 by any prime $p$ with $1 < p \leq \sqrt{n}$.
 
+<details>
+<summary>Proof</summary>
+
 Proving $\implies$
 This follows directly from the definition of prime.
 
@@ -29,43 +34,63 @@ Clearly if there were such a prime $p$ in range $(1, \sqrt{n}]$, n would not be
 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.
+</details>
 
-### Facto
+### 2.2
 
 Prove that if $n > 4$ is composite then $n | (n - 1)!$.
 
+<details>
+<summary>Proof</summary>
+
 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 $(n-1)!$.
-
-### Another Fact
+Hence, all prime factors of $n$ can be accounted for in $(n-1)!$ when $n > 4$, so $n$ is a factor of $(n-1)!$
+</details>
 
+### 2.3
 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.
 
-Hint: Compute the remainder upon division by 4 of each of $(4m)^2$, $(4m + 1)^2$, $(4m + 2)^2$, and $(4m + 3)^2$.
+<details>
+<summary>Hint</summary>
+
+Compute the remainder upon division by 4 of each of $(4m)^2$, $(4m + 1)^2$, $(4m + 2)^2$, and $(4m + 3)^2$.
+</details>
+
+<details>
+<summary>Proof</summary>
 
 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 $4k+3$.
+</details>
+
+### 2.4
 
-### Follow-up Fact
+(Follow-up to 2.3)
 
-Prove that no integer in the sequence
+Prove that no integer in the sequence below is a perfect square.
 
 $11$, $111$, $1111$, $11111$, $\dots$
 
-is a perfect square.
+<details>
+<summary>Hint</summary>
+
+$111 \dots 111 = 111 \dots 108 + 3 = 4k+ 3$.
+</details>
 
-Hint: $111 \dots 111 = 111 \dots 108 + 3 = 4k+ 3$.
+<details>
+<summary>Proof</summary>
 
-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.
+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.
+</details>
 
-### A series of fun facts
+### 2.5
 
 #### 3
 
@@ -77,51 +102,76 @@ Proof: write $n = a + 10b + 100c + \dots$, where the digits of $n$ are $a, b, c,
 
 A number $n \in \mathbb{Z}$ is divisible by 5 if and only if the rightmost digit is 5 or 0.
 
+<details>
+<summary>Proof</summary>
+
 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.
+</details>
 
 #### 9
 
 A number $n \in \mathbb{Z}$ is divisible by 9 if and only if the sum of the digits is.
 
+<details>
+<summary>Proof</summary>
+
 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.
+</details>
 
 #### 11
 
 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 $0 \pmod{11}$.
 
+<details>
+<summary>Proof</summary>
+
 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.
 
-Additional note:
+##### Additional Note
+
 *Definition* (Palindrome): let the digits of $n \in \pmod{11}$ be of the form $abc...cba$. Then $n$ is a palindrome.
 
 *Remark*: From the proof above, it follows that all even-length palindromes are divisible by 11.
 
 *Conjecture*: All palindromes divisible by 11 are such that the product of their digits is a perfect square.
+</details>
 
-### One more
+### 2.6
 
 Prove that if $p$ is a positive integer such that both $p$ and $p^2 + 2$ are prime, then $p = 3$.
 
+<details>
+<summary>Proof</summary>
+
 Trivially, 2 does not satisfy the conditons and 3 does.
-Now we consider only prime $p > 3$.
+Now we consider only primes $p > 3$.
 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.
+</details>
 
 ## Hard
 
 ### Fermat's Last Theorem
 
-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 $2$.
+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 $2$.
+
+<details>
+<summary>Proof</summary>
+
+I am unable to find a copy of the original proof by Andrew Wiles, but [this source](https://people.math.wisc.edu/~nboston/869.pdf) should be pretty close. Yes, the proof is that long.
+
+Read more about the theorem and its incredible story [here](https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem).
+</details>
 
 ### Collatz Conjecture
 
@@ -138,6 +188,18 @@ $
 
 Prove that for all $n$, this function will eventually return 1.
 
+<details>
+<summary>Proof</summary>
+
+There is no known proof of this conjecture. Read more about it [here](https://en.wikipedia.org/wiki/Collatz_conjecture).
+</details>
+
 ### Goldbach's Conjecture
 
 Prove that every even $n \in \mathbb{n}$ greater than 2 is the sum of two primes.
+
+<details>
+<summary>Proof</summary>
+
+There is no known proof to this conjecture. Read more about it [here](https://en.wikipedia.org/wiki/Goldbach%27s_conjecture).
+</details>