Question

Difficulty: MediumOdd and Even Integers (Parity)

For any integer nn, which of the following expressions MUST be an even integer?

  1. (n+1)3(n1)3(n + 1)^3 - (n - 1)^3Answer
  2. B
    n3n+1n^3 - n + 1
  3. C
    (n+1)2+n2(n + 1)^2 + n^2
  4. D
    3n2+3n+13n^2 + 3n + 1
  5. E
    n2+n3n^2 + n - 3

Answer

The expression (n+1)3(n1)3(n + 1)^3 - (n - 1)^3 MUST be an even integer for all integer values of nn.
Simplifying the expression (n+1)3(n1)3(n + 1)^3 - (n - 1)^3 yields 6n2+26n^2 + 2, which equals 2(3n2+1)2(3n^2 + 1). Since 3n2+13n^2 + 1 is always an integer for any integer nn, multiplying by 2 guarantees the expression is an even integer for all values of nn, including zero and negative integers.

Step-by-Step Solution

1
Expand both cubic terms algebraically.
(n+1)3=n3+3n2+3n+1(n + 1)^3 = n^3 + 3n^2 + 3n + 1 and (n1)3=n33n2+3n1(n - 1)^3 = n^3 - 3n^2 + 3n - 1.
Expanding the terms allows combining like terms to simplify the expression.
2
Subtract the expanded expression (n1)3(n - 1)^3 from (n+1)3(n + 1)^3.
(n3+3n2+3n+1)(n33n2+3n1)=6n2+2(n^3 + 3n^2 + 3n + 1) - (n^3 - 3n^2 + 3n - 1) = 6n^2 + 2.
The cubic terms (n3n^3) and linear terms (3n3n) cancel out.
3
Factor out a common factor of 2 from the simplified expression.
6n2+2=2(3n2+1)6n^2 + 2 = 2(3n^2 + 1).
Any integer that can be expressed as 2k2k, where kk is an integer, is by definition even. Since nn is an integer, 3n2+13n^2 + 1 is an integer, making 2(3n2+1)2(3n^2 + 1) an even integer for all nn.

Key Concept

Parity rules for algebraic expressions and consecutive integer products

Alternative Method

Instead of algebraic expansion, test values of nn with different parities, including n=0n = 0 (even) and n=1n = 1 (odd). For n=0n = 0, (0+1)3(01)3=1(1)=2(0 + 1)^3 - (0 - 1)^3 = 1 - (-1) = 2 (even). For n=1n = 1, (1+1)3(11)3=80=8(1 + 1)^3 - (1 - 1)^3 = 8 - 0 = 8 (even). Because n+1n + 1 and n1n - 1 always have the same parity, their cubes also share the same parity, so their difference is always even.
Estimated Time:1m 30s
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