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    Number Theory

    Let $\displaystyle a,b \in \mathbb{Z} .$ with (a,4)=2 and (b,4)=2. find (a+b, 4) and prove that your answer is correct.
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    Quote Originally Posted by JCIR View Post
    Let $\displaystyle a,b \in \mathbb{Z} .$ with (a,4)=2 and (b,4)=2. find (a+b, 4) and prove that your answer is correct.
    $\displaystyle (a + b,\,4) = 4$

    $\displaystyle \emph{Proof: }$ First, note that the only possible divisors of 4 are 1, 2, and 4, so we only need show that $\displaystyle 4\mid(a + b)$

    $\displaystyle (a,\,4) = 2\text{ and }(b,\,4)=2\Rightarrow2\mid a\text{ and }2\mid b\Rightarrow\exists p,\,q\in\mathbb{Z},\;a = 2p,\,b = 2q$

    $\displaystyle \Rightarrow a + b = 2(p + q)$

    But $\displaystyle 2\nmid p$, for if so, $\displaystyle \exists m\in\mathbb{Z},\;a = 2(2m) = 4m\Rightarrow4\mid a$ and so $\displaystyle (a,\,4) = 4$ and not 2 as required. Similarly, $\displaystyle 2\nmid q$.

    Thus $\displaystyle p$ and $\displaystyle q$ are odd so their sum $\displaystyle p + q$ must be even (i.e., $\displaystyle \exists s,\,t\in\mathbb{Z},\;p = 2s + 1\text{ and }q = 2t + 1$ $\displaystyle \Rightarrow p + q = 2s + 2t + 2 = 2(s + t + 1)$ with $\displaystyle s + t + 1\in\mathbb{Z}$). So $\displaystyle 4\mid2(p + q)\Rightarrow4\mid(a + b)$ as required to show that $\displaystyle (a + b,\,4) = 4\quad\square$
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