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Prove if $k$ is an integer, then $k^2 - 3k$ is an even integer

Im having some trouble with this proof. Im using proof by contrapositive (if $k^2 - 3k$ is odd integer then $k$ is not an integer), and so I set $k^2 - 3k = 2k + 1$ and rearranged it to get $k^2 -5k -1 = 0$, but I don't know what to do from here, how can I show that $k$ is not an integer?

Parcly Taxel
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user677216
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    Consider $k(k-3)$ . – Randall May 29 '19 at 01:38
  • Your step from $k^2-3k$ being odd to it being $2k+1$ is unwarranted. Maybe you meant to write something like $k^2-3k=2m+1$ for some other integer $m$? – kimchi lover May 29 '19 at 01:40
  • Integers are either of the form 2k+1 or 2m for integers m,k. Consider the cases when k is odd and when k is even. –  May 29 '19 at 01:45

3 Answers3

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$k^2$ and $3k$ are either both even or both odd, hence their difference is even.

Rob Arthan
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You have made a mistake in assigning the same variable $k$ for the odd number on the RHS of $k^2-3k=2k+1$; it should be some other letter, e.g. $k^2-3k=2l+1$.

A simpler approach would be factoring $k^2-3k=k(k-3)$. $k$, being an integer, is even or odd, and this means $k$ or $k-3$ is even respectively, so the product must always be even.

Parcly Taxel
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  • yea I was thinking that the RHS needed a different variable as well. however I don't intuitively understand why the variables need to be different – user677216 May 29 '19 at 01:46
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    @user677216 If you write $k$ in the RHS, you have seen yourself that it leads to a quadratic in $k$, which means $k$ can have only two possible values. But $k$ is free at the beginning; it can have infinitely many possible values. This is absurd. – Parcly Taxel May 29 '19 at 01:47
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If $k$ is even, $k^2$ is even and $3k$ is even, so $k^2-3k$ is even.

If$k$ is odd so are $k^2$ and $3k$. $k^2-3k$ is the difference of two odd numbers, so it is even.

herb steinberg
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