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If integers a and b are expressible in the form $4k+1$ where k is a integer then $ab$ is also expressible in the form.

This seems like a simple example But I am wondering if I did it correct.

Since $a=4k+1$ and $b=4k+1$

Then

$a(b)=(4k+1)(4k+1)$

$=16k^2+4k+4k+1=16k^2+8k+1$

$=(4k+1)^2=(a)(b)$

Fernando Martinez
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    Note: this statement is incredibly trivial if you use modular arithmetic. $a=4k+1$ for some integer $k$ means that $a\equiv 1~\text{mod}4$, and similarly $b=4l+1$ for some integer $l$ means that $b\equiv 1~\text{mod}4$. So, $ab \equiv 1\cdot 1 \equiv 1~\text{mod}4$, implying that $ab=4n+1$ for some integer $n$. – JMoravitz Sep 12 '15 at 18:04
  • Also the direct way multiplying $(4k+1)(4l+1)$ is not incredibly nontrivial. – Dietrich Burde Sep 12 '15 at 18:33

1 Answers1

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What you have to show is that with $a=4k+1$ and $b=4l+1$ also the product $ab$ is congruent $1$ modulo $4$. This follows from $$ a\cdot b=(4k+1)(4l+1)=16kl+4k+4l+1=4(4kl+k+l)+1. $$

Dietrich Burde
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