Prove that for any odd integer m, the number (m4 + 4m2 + 11)/16 is an integer.

Step 1: Let m be an odd integer. We can express any odd integer m in the form m = 2k+1 for some integer k. Step 2: Find expressions for m^2 and m^4 in terms of k. m^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1 Since k and k+1 are consecutive integers, one of them must be even. Therefore, k(k+1) is always an even integer. Let k(k+1) = 2j for some integer j. Then, m^2 = 4(2j) + 1 = 8j+1. Now, find m^4: m^4 = (m^2)^2 = (8j+1)^2 = (8j)^2 + 2(8j)(1) + 1^2 = 64j^2 + 16j + 1 Step 3: Substitute these expressions into the given number. The number is (m^4 + 4m^2 + 11)/(16). Substitute m^4 = 64j^2 + 16j + 1 and m^2 = 8j+1: ((64j^2 + 16j + 1) + 4(8j+1) + 11)/(16) (64j^2 + 16j + 1 + 32j + 4 + 11)/(16) (64j^2 + (16j + 32j) + (1 + 4 + 11))/(16) (64j^2 + 48j + 16)/(16) Step 4: Simplify the expression. Factor out 16 from the numerator: (16(4j^2 + 3j + 1))/(16) 4j^2 + 3j + 1 Since j is an integer, 4j^2 + 3j + 1 is also an integer. Therefore, the given number is an integer. The number is an integer. Step 1: Combine the square roots using the property sqrt(x) × sqrt(y) = sqrt(xy). sqrt((b)/(a)) × sqrt((c)/(b)) × sqrt((a)/(c)) = sqrt((b)/(a) × (c)/(b) × (a)/(c)) Step 2: Multiply the fractions inside the square root. sqrt((b · c · a)/(a · b · c)) Step 3: Cancel out common terms in the numerator and denominator. sqrt((abc)/(abc)) = sqrt(1) Step 4: Evaluate the square root. sqrt(1) = 1 Thus, sqrt((b)/(a)) × sqrt((c)/(b)) × sqrt((a)/(c)) = 1. sqrt((b)/(a)) × sqrt((c)/(b)) × sqrt((a)/(c)) = 1 Step 1: The problem states that sqrt(n) = (3)/(2). Step 2: We need to find the value of -sqrt(n). Substitute the given value of sqrt(n) into the expression: -sqrt(n) = -((3)/(2)) -sqrt(n) = -(3)/(2) -(3)/(2) Step 1: Start with the known inequality that the square of any real number is non-negative. For any real numbers a and b, (a-b)^2 0. Step 2: Expand the inequality. a^2 - 2ab + b^2 0 Step 3: Since a and b are positive real numbers, their product ab is also positive. We can divide the entire inequality by ab without changing the direction of the inequality sign. (a^2 - 2ab + b^2)/(ab) (0)/(ab) (a^2)/(ab) - (2ab)/(ab) + (b^2)/(ab) 0 Step 4: Simplify the terms. (a)/(b) - 2 + (b)/(a) 0 Step 5: Add 2 to both sides of the inequality. (a)/(b) + (b)/(a) 2 This proves the statement. (a)/(b) + (b)/(a) 2 Step 1: Start with the known inequality that the square of any real number is non-negative. Since x is a real number and x 0, x - (1)/(x) is a real number. Therefore, (x - (1)/(x))^2 0. Step 2: Expand the square. (x - (1)/(x))^2 = x^2 - 2 · x · (1)/(x) + ((1)/(x))^2 x^2 - 2 + (1)/(x^2) Step 3: Substitute the expanded form back into the inequality. x^2 - 2 + (1)/(x^2) 0 Step 4: Add 2 to both sides of the inequality. x^2 + (1)/(x^2) 2 This proves the statement. x^2 + (1)/(x^2) 2