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Assume that $(4k + 3) ^ 2 - (4k + 3)$ is not divisible by 4. If this is true, prove that $(4(k+1) + 3) ^ 2 - (4(k+1) + 3)$ is not divisible by 4.

I need to prove this for my induction problem, and I've tried manipulating the second expression as much as I can but haven't arrived at a situation where I can use the first expression. Any help is appreciated, could anyone show me how to do this? Thank you.

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Let $a=4k+3$. Then $$b=(4(k+1) + 3) ^ 2 - (4(k+1) + 3)=(a+4)^2-(a+4)=(a^2-a)+4(2a+3)$$ So, $b$ is a multiple of $4$ iff $a^2-a$ is a multiple of $4$, which it is not, by hypothesis.

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  • $\begingroup$ Thank you very much, I just have one question, could you explain how (a + 4)^2 - (a + 4) become (a^2 - a) + 8a, please? $\endgroup$ – Daniel Cook Oct 2 '13 at 3:21
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    $\begingroup$ My mistake. I've edited my answer. $\endgroup$ – lhf Oct 2 '13 at 3:23
  • $\begingroup$ Thank you so much! I finally understand how to manipulate it this way. $\endgroup$ – Daniel Cook Oct 2 '13 at 3:26

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