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Difficulty: MediumOdd and Even Integers (Parity)

If mm and nn are integers such that m2n+mm^2 n + m is an odd integer, which of the following expressions MUST be an even integer?

  1. A
    m2+nm^2 + n
  2. B
    m+2nm + 2n
  3. C
    m(n+1)m(n + 1)
  4. m2+n2+1m^2 + n^2 + 1Answer
  5. E
    3m−n3m - n

Answer

The expression m2+n2+1m^2 + n^2 + 1 must be an even integer.
Factoring the given expression gives m(mn+1)=oddm(mn + 1) = \text{odd}, which implies mm is odd and mn+1mn + 1 is odd. Consequently, mnmn is even, which forces nn to be even (since mm is odd). Substituting m=oddm = \text{odd} and n=evenn = \text{even} into the expression m2+n2+1m^2 + n^2 + 1 yields odd+even+1=even\text{odd} + \text{even} + 1 = \text{even}. Thus, this expression must be an even integer.

Step-by-Step Solution

1
Factor the given expression to analyze its parity.
m2n+m=m(mn+1)m^2 n + m = m(mn + 1).
Factoring out mm isolates the product of two factors.
2
Determine the parity of each factor.
Since m(mn+1)m(mn + 1) is odd, both mm and (mn+1)(mn + 1) must be odd integers.
The product of two integers is odd if and only if both factors are odd.
3
Determine the parity of nn.
Since mn+1mn + 1 is odd, mnmn must be even. Because mm is odd, nn must be even.
For the product mnmn to be even when mm is odd, nn must be an even integer (including 0).
4
Evaluate the target expression m2+n2+1m^2 + n^2 + 1.
m2m^2 is odd, n2n^2 is even, so m2+n2+1=odd+even+odd=evenm^2 + n^2 + 1 = \text{odd} + \text{even} + \text{odd} = \text{even}.
Summing two odd integers and one even integer always results in an even integer.

Key Concept

Parity properties under addition, subtraction, and multiplication
Estimated Time:1m 30s
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Topics

Odd and Even Integers (Parity)

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