Find prime p, q, s.t. p^2

Find prime p, q, s.t. p^2 - 2q^2 = 1

Problem:
Find prime \(p, q\) which satisfies \(p^2 - 2q^2 = 1\).

Solution:
We see that \(p = 3, q = 2\) is a solution. We want to show that this is the only solution.

Suppose there exists a solution with \( q > 2 \). Rearranging the equation we have \( (p-1)(p+1) = 2q^2 \). Since RHS is divisible by 2, we know that \(p\) must be an odd prime. Hence, both \( p-1 \) and \( p+1 \) are divisible by 2. That means LHS is at least divisible by 4. Hence \( 2q^2 \) must be divisible by 4. This is only possible when \( 2 | q \). Contradiction.

QED

Comments

641 | 2^32 + 1

Problem: Show that \( 641 | 2^{32} + 1 \). Solution: (From Problem Solving Strategies, Arthur Engel) \( 641 = 625 + 16 = 5^4 + 2^4 \). So \( 641 | 2^{32} + 2^{28} \cdot 5^4 \). Also, \( 641 = 640 + 1 = 2^7 \cdot 5 + 1\). So \( 641 | (2^7 \cdot 5)^4 - 1 = 2^{28}\cdot 5^4 - 1 \). Hence \( 641 | 2^{32} + 2^{28} \cdot 5^4 -(2^{28}\cdot 5^4 - 1) \). QED

Derive e

Prove that \( \left( 1 + \frac{1}{n} \right)^n \) converges. Solution Let \( G(n) = \left( 1 + \frac{1}{n} \right)^n \). Binomial expansion, \( G(n) = \sum_{k = 0}^{n} \frac{ \binom{n}{k} }{n^k} \). By simple inequality argument, \( \frac{ \binom{n}{k} }{n^k} < \frac{1}{k!} \). So if we let \( F(n) = \sum_{k = 0}^{n} \frac{1}{k!} \), we have \( G(n) < F(n) \). By simple inequality argument, \( F(n) < 1 + \sum_{k = 1}^{n} \frac{1}{2^{k-1}} < 3 \). So G and F are bounded above. By AM-GM, \( \frac{G(n-1)}{G(n)} = \left( \frac{ n^2 }{ n^2 - 1} \right)^{(n-1)} \frac{ n}{n+1} < \left( \frac{ \frac{ n^2 }{ n^2 - 1} (n-1) + \frac{ n}{n+1} }{n} \right)^n = 1 \). So \( G(n-1) < G(n) \). Since G is monotonic increasing and bounded above, G converges. Similarly F converges. Let's also prove that G and F converges to the same limit. By studying individual terms in \( F(n) - G(n) \) and making simple inequality arguments, we can show that for each \( \epsilon > 0 \) w...

Random walk by the cliff

You are one step away from falling down a cliff. In each step, you have 1/3 chance of moving towards the cliff, 2/3 chance of moving away from the cliff. What is the probability you fall down the cliff? Solution It is 1/2, and there are 3 ways to approach this. Let's say we are at position 0, and the cliff is at position -1. We go +1 with probability p, and go -1 with probability 1-p. Approach 1. Catalan number: All paths that end with falling down the cliff can be broken down to two parts: 0 -> ... -> 0 and 0 -> -1. The path 0 -> ... -> 0 is Catalan; if we fix the path length to 2n, then the number of paths would be C_n. What we want to compute is the total probability = (1-p) sum { C_n (p(1-p))^n }. We can use the generating function for Catalan to arrive at: if p <= 1/2, probability = 1; if p > 1/2, probability = 1/p - 1. Approach 2. Let P(i) be the probability of reaching -1 from position i. P(0) = 1-p + pP(1) Also, P(1) can be derived as follows: can we r...

Maths by Chance

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