Maths by Chance

I was damned confused! I think I understand it now, so I'd like to share how to resolve this "paradox". First, to recap, suppose you have a process / experiment that produces a confidence interval [x, y], for which you say, if we repeat this experiment a lot of time, 95% of the cases we'll see that the [x, y] covers something that we are estimating, e.g. the true mean. We call this 95% Confidence Interval (CI) for the true mean. But then, your instructor, or the author of the book you are reading, will immediately follow up by the statement "Be careful! It's wrong to say that the CI you computed has 95% chance of containing the true mean". Confusion ensues. You lose sleep, you don't eat as much. Your parents, friends, and loved ones start to get worried because you don't look so well. You, on the other hand, has lost all confidence in your understanding of statistics and basics of probabilities. WHY on earth?? If 95% of the CIs produced by the re...

Problem: Among 5 integers, there are always 3 with sum divisible by 3. (From Problem Solving Strategies, Arthur Engel) Solution: Proof by Pigeon Hole Principle. An integer is either 0, 1 or 2 \( \pmod 3 \). Imagine placing 5 integers into those 3 boxes. If we have at least one in each, then we can pick one from each, with sum divisible by 3. Otherwise, we'll have at least 3 integers in one of the boxes. Pick those 3. QED

Problem: Show that if \( p \) and \( p^2 +2 \) are prime, then so is \( p^3 + 2 \). Solution: \(p\) can't be 2. For \(p = 3 \), the statement is true. If \(p > 3\), we know that \(p^2 = 1 \pmod 3 \). Hence \(p^2 + 2 \) is always divisible by 3. QED.

Problem: If \(n\) is even, then \( 323 | 20^n + 16^n - 3^n - 1 \). Solution: Let \( n = 2k \). We have \( 17 | 400^k - 9^k \) and \( 17 | 256^k - 1 \). Hence \( 17 | 20^n + 16^n - 3^n - 1 \). Also \( 19 | 20^n -1 \) and \( 19 | 256^k - 9^k \). Hence  \(19 | 20^n + 16^n - 3^n - 1 \). Hence it is divisible by \( 17 \times 19 = 323 \). QED

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