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If $n$ is a positive integer, then $n^2+n$ is even.

$n=2k$ for some k in the integers

$n^2 +n$ = $4k^2 + 2k$ = $2(2k^2+k)$

as $2k^2+k$ is an integer $n^2+n$ is even.

Jst wondering if this proof is ok. AM i allowed to say $2k^2+k$ is an integer? Thanks

Bill Dubuque
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Carlos Bacca
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  • "AM I allowed to say $2k^2+ k$ is an integer?" Yes, of course The set of integers is "closed" under both addition and multiplication. But it is much simpler say that $n^2+ n= n(n+ 1)$. Do you see that one of n or n+1 must be even so that the product is even. – user247327 Nov 06 '18 at 18:11
  • The main parts are fine but I would start by saying: “suppose n is even. Then n=2k...” as the answer below indicates you also have to consider what happens if n is odd. And yes, $2k^2 +k$ Is an integer by the integer axioms, so you can just state it. – JavaMan Nov 06 '18 at 18:11
  • @user247327 Yeah that is easier isnt it – Carlos Bacca Nov 06 '18 at 18:17

3 Answers3

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You have to consider a case where n is odd... n = 2k + 1

Peter Chikov
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You have set $n=2k$ for some k in the integers, so your proof is ok for even $n$.

Now for odd $n$, you can do the same:

set $n=2k+1$ for some k

$n^2 +n$ = $4k^2 + 4k + 2k + 2$ = $2(2k^2+3k + 1)$

so this is even.

Andreas
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  • One also may need to prove that every integer is odd or even, which follows from the division algorithm (divide by $2$ and examine the remainder). – Bill Dubuque Nov 06 '18 at 18:13
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When you say $n = 2k$ for some integer $k$, you're assuming that $n$ is even. That was not part of the given assumptions.

But you have successfully shown that if $n$ is an even integer, then $n^2 + n$ is also even (yes, $2k^2 + k$ is an integer: if you start with an integer, then no amount of multiplication and addition will make it stop being an integer).

You have to check that it also holds for $n$ odd, i.e. $n = 2k+1$.

Arthur
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