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I have to prove that $(4k + 3) ^2 - (4k + 3)$ is not divisible by $4$.

What would be the best approach for this, proof by induction or contradiction?

I've tried both and haven't got very far. Any hints would be appreciated, I'm not looking for a full answer..I wanna try it out myself but I need some help on where to begin.

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  • 2
    $\begingroup$ Have you learned about modular arithmetic? $\endgroup$ – Tobias Kildetoft Sep 30 '13 at 19:09
  • $\begingroup$ Not really no.. $\endgroup$ – Daniel Cook Sep 30 '13 at 19:11
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    $\begingroup$ Neither contradiction nor induction: calculation. $\endgroup$ – André Nicolas Sep 30 '13 at 20:14
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The polynomial is simple enough that it’s no problem simply to multiply it out:

$$\begin{align*} (4k+3)^2-(4k+3)&=16k^2+24k+9-4k-3\\ &=16k^2+20k+6\\ &=4\left(4k^2+5k+1\right)+2\;, \end{align*}$$

which is clearly not a multiple of $4$. This is perhaps a little less elegant than njguliyev’s solution, but it works just fine.

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  • $\begingroup$ I can't read the formatted text for some reason :/ $\endgroup$ – Daniel Cook Sep 30 '13 at 19:45
  • $\begingroup$ @extremez: Very strange; I didn’t do anything unusual. What browser are you using? In any case, I just multiplied out and collected terms to get $4\left(4k^2+5k+1\right)+2$. $\endgroup$ – Brian M. Scott Sep 30 '13 at 19:47
  • $\begingroup$ No it's not just your answer, I haven't been able to read formatted text in any answers or questions or even comments period. i'm using the latest Google Chrome and it hasn't worked on Firefox either. it used to work before so it's weird that it doesn't anymore, thank you for clarifying that anyways, it's very helpful. $\endgroup$ – Daniel Cook Sep 30 '13 at 19:49
  • $\begingroup$ @extremez: You’re welcome. If the problem persists, you might ask about it on meta. $\endgroup$ – Brian M. Scott Sep 30 '13 at 19:57
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$(4k + 3)^2 - (4k + 3) = (4k + 3)(4k + 3-1) = (4k + 3)(4k + 2)=2(4k + 3)(2k + 1)$.

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As

$$(4k+3)^2-(4k+3)=(4k+3)\{(4k+3)-1\}=(4k+3)(4k+2)$$

Now, $4k+3=2(2k+1)+1$ is odd

and

$4k+2=2(2k+1)\not\equiv0\pmod4$

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