Nice proof by induction that symmetric matrix is orthogonally diagonalizable

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

sum of 3 out of 5 is divisible by 3

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

2n+1 3n+1 squares then 5n+3 not prime

Problem: Prove that if \( 2n+1 \) and \( 3n+1 \) are both squares, then \( 5n+3 \) is not a prime. Solution: Proof by contradiction. Suppose that there is n such that 5n+3 is prime. Note that \( 4(2n+1) - (3n+1) = 5n+3 \). Let \( 2n+1 = p^2 \) and \( 3n+1 = q^2 \), where \( p, q > 0 \). We have \( (2p-q)(2p+q) = 5n+3 \). Since RHS is a prime, we must have \(2p - q = 1 \) and \( 2p + q = 5n+3 \). Solving for \( q \) we get \( 2q = 5n + 2 \). Substituting, we get \( 2q = q^2 + 2n + 1 \), or \( -2n = (q-1)^2 \). Since RHS is \( \geq 0 \), we can only have equality when \( n= 0 \) and \( q = 1 \). In this case, we have \( 5n + 3 = 8 \) not a prime. QED.

Maths by Chance

Search This Blog