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1. Use the theory of congruence to find the remainder of 3100 (a) when it is divided by 28; and (b) when it is divided by 29. 2. Prove that for any integer n, n³ = 0, 1,8 (mod9). 3. Use mathematical induction to prove the following statements. (a) For all n € N\{0}, -1² +2²-3² +4² + (−1)"n² = (-1)" _ "(n+1) 2 (b) Suppose a 1. Then for all n N\{0}. 11 Σra-1- r=1 1-(n+1)a" + na"+1 (1-a)² (e) For any neN\{0), 7/2+1+32-1. (d) For any n € N, (√3+1)2n+1-(√3-1)2+1 is an integer which is divisible by 2+1 (e) For any ne N\{0}. (g) Let f(x) = . Then for any n € N. 11 1 Σ≤2- r=1 (f) Let a₁ = 1 and 0₂ = 1. Define anan-1 + 0-2 when n>3. Then for all n € N\{0}, Σ ar=an+2 -1. ral (4 points) (6 points each) (1+x²) f(+2)(x) + 2(n+2)rf(+¹)(x) + (n +2)(n+1)f)(x) = 0. 4. Let an be a sequence of positive numbers that satisfies an (4 points) (6 points) a₁ + a₂ + a3 + + a₁ = for n E N\{0}. Prove that a, = 2n-1 for all n € N\{0}. (6 points) (6 points) 5. For any n € N. (√3+1)2+1-(√3-1)2+1 is an integer which is divisible by 2+1 6. The goal of this problem is to prove that for every positive integer n, at least one of n, n+1,n+2, n+3 can be written as a sum of distinct square numbers (a square number is a number which is a square of an integer). We will prove this statement in two parts. (a) Show that 1, 4, 5, 9, 10, 13, 14, 16 can be written as a sum of distinct square numbers. Hence, the statement is true for positive integers n ≤ 16. (2 points) (b) Using strong induction, or otherwise, show that the statement is true for integers n 2 16. Hint: We need to bound n by the largest square number that is less than or equal to n, meaning k²