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Nice proof by induction that symmetric matrix is orthogonally diagonalizable

https://people.math.carleton.ca/~kcheung/math/notes/MATH1107/wk10/10_orthogonal_diagonalization_proof.html

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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
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Problem: Find the n-th non-square number. (Adapted from Problem Solving Strategies, Arthur Engel ) If you have CS background, I invite you to come up with an O(log N) solution. Done? Good. Below I present the O(1) approach. Solution: The n-th non-square number is given by \( F(n) = \left \lfloor n + \sqrt{n} + 1/2  \right \rfloor \). Proof: Let G(n) be the same as F(n) except for the floor function. Investigate \( m \neq F(n) \). There is the largest \( n \) s.t. \( F(n) < m \). Hence \( m+1 \leq F(n+1) \). We conclude that \(G(n) < m \) and \(m+1 \leq G(n+1) \). Rearranging the terms and squaring the inequalities, we get: \( n < (m-n-1/2)^2 \) and \( (m-n-1/2)^2 < n+1 \), or \( n - 1/4 < (m-n)^2 - (m-n) < n + 3/4 \). Since the middle term is an integer, we see that \( (m-n)^2 - (m-n) = n \). Simplifying, we get \(m = (m-n)^2\), showing that m is a square. In other words, F(n) skips all squares. Now, we will show that F(n) is the n-th non...
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