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Let m be an odd integer. Show that the number racm4 + 4m2 + 1116 is an integer.

Mathematics

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1. Let mm be an odd integer. Show that the number m4+4m2+1116\frac{m^4 + 4m^2 + 11}{16} is an integer.

  1. Express mm in terms of an integer kk.

    Since mm is an odd integer, we can write m=2k+1m = 2k + 1 for some integer kk.

  2. Find an expression for m2m^2.

    m2=(2k+1)2=4k2+4k+1=4k(k+1)+1m^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1

    Since k(k+1)k(k+1) is always an even integer (the product of two consecutive integers), let k(k+1)=2jk(k+1) = 2j for some integer jj.
    Then, m2=4(2j)+1=8j+1m^2 = 4(2j) + 1 = 8j + 1.

  3. Substitute m2=8j+1m^2 = 8j+1 into the given expression's numerator.

    The numerator is m4+4m2+11m^4 + 4m^2 + 11.

    m4+4m2+11=(m2)2+4m2+11m^4 + 4m^2 + 11 = (m^2)^2 + 4m^2 + 11 =(8j+1)2+4(8j+1)+11= (8j+1)^2 + 4(8j+1) + 11 =(64j2+16j+1)+(32j+4)+11= (64j^2 + 16j + 1) + (32j + 4) + 11 =64j2+16j+32j+1+4+11= 64j^2 + 16j + 32j + 1 + 4 + 11 =64j2+48j+16= 64j^2 + 48j + 16
  4. Factor out 16 from the numerator.

    64j2+48j+16=16(4j2+3j+1)64j^2 + 48j + 16 = 16(4j^2 + 3j + 1)
  5. Substitute the factored numerator back into the original expression.

    m4+4m2+1116=16(4j2+3j+1)16\frac{m^4 + 4m^2 + 11}{16} = \frac{16(4j^2 + 3j + 1)}{16} =4j2+3j+1= 4j^2 + 3j + 1

    Since jj is an integer, 4j2+3j+14j^2 + 3j + 1 is also an integer.
    Therefore, the number m4+4m2+1116\frac{m^4 + 4m^2 + 11}{16} is an integer.

    2. If a,b,a, b, and cc are positive real numbers, then show that: ba⋅cb⋅ac=1\sqrt{\frac{b}{a}} \cdot \sqrt{\frac{c}{b}} \cdot \sqrt{\frac{a}{c}} = 1

  6. Combine the terms under a single square root.

    Since a,b,ca, b, c are positive real numbers, we can use the property x⋅y⋅z=xyz\sqrt{x} \cdot \sqrt{y} \cdot \sqrt{z} = \sqrt{xyz}.

    ba⋅cb⋅ac=ba⋅cb⋅ac\sqrt{\frac{b}{a}} \cdot \sqrt{\frac{c}{b}} \cdot \sqrt{\frac{a}{c}} = \sqrt{\frac{b}{a} \cdot \frac{c}{b} \cdot \frac{a}{c}}
  7. Simplify the expression inside the square root.

    b⋅c⋅aa⋅b⋅c\sqrt{\frac{b \cdot c \cdot a}{a \cdot b \cdot c}}

    Cancel out the common terms a,b,ca, b, c from the numerator and denominator.

    =1= \sqrt{1}
  8. Evaluate the square root.

    1=1\sqrt{1} = 1

    Thus, ba⋅cb⋅ac=1\boxed{\sqrt{\frac{b}{a}} \cdot \sqrt{\frac{c}{b}} \cdot \sqrt{\frac{a}{c}} = 1}.

    3. Suppose n=32\sqrt{n} = \frac{3}{2}, find −n-\sqrt{n}

  9. Substitute the given value of n\sqrt{n}.

    We are given that n=32\sqrt{n} = \frac{3}{2}.
    We need to find the value of −n-\sqrt{n}.

    −n=−(32)-\sqrt{n} = - \left(\frac{3}{2}\right)
  10. Simplify the expression.

    −n=−32-\sqrt{n} = -\frac{3}{2}

    The value of −n-\sqrt{n} is −32\boxed{-\frac{3}{2}}.

    4. Let aa and bb be positive real numbers. Show that ab+ba≥2\frac{a}{b} + \frac{b}{a} \ge 2

  11. Apply the AM-GM inequality.

    For any two positive real numbers xx and yy, the Arithmetic Mean-Geometric Mean (AM-GM) inequality states that $\

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