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I am helping my sister study for the praxis exam of this study book, and I reviewed a question based on number theory. I see it involves constant integers my question is:

If $m$ and $n$ are consecutive integers, which can never be even? Choose all that apply.

However, I am focusing on this particular one, \begin{equation}n(m+1)^2 \end{equation} My question is based on these two substitutions which tell me that this equation is odd: \begin{equation}1(2+1)^2 =9\end{equation}\begin{equation}3(2+1)^2=27 \end{equation} These are odd. However, in the back of the book it says that this can be even are my substitutions wrong.

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Jose M Serra
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1 Answers1

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The expression $n(m+1)^2$ being even or odd depends on $n$: $$n \text{ even}\Longrightarrow n(m+1)^2\text{ even}$$ $$n \text{ odd}\Longrightarrow n(m+1)^2\text{ odd}$$ That's because $n$ and $m$ are consecutive integers, so $n$ even implies $m+1$ even (analogous for odd). Seeing that

  • The multiplication of two even integers is another even (same for two odds)
  • The square of an even integer is another even (same for two odds) (This one's a corolary of the statement above).

(this statements have an easy proof using prime factors) you can conclude the implications a gave at first.

Alejandro Bergasa Alonso
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