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Question Number 88261 by ar247 last updated on 09/Apr/20

Answered by Joel578 last updated on 09/Apr/20

$$u\equiv v\left(\mathrm{mod}m\right)\Rightarrow u-v=mx,x\in \mathbb{Z}$$$$\left(1\right)u=mx+v$$$$\mathrm{Since}\mathrm{gcf}(v,m)\mathrm{divides}\mathrm{both}v\mathrm{and}\mathrm{m},\mathrm{it}\mathrm{also}\mathrm{divides}u,$$$$\mathrm{hence}\mathrm{gcf}(v,m)\mathrm{divides}\mathrm{gcf}(u,m)$$$$\Rightarrow \mathrm{gcf}(v,m)\mid \mathrm{gcf}(u,m)$$$$$$$$\left(2\right)v=u-mx$$$$\mathrm{Since}\mathrm{gcf}(u,m)\mathrm{divides}\mathrm{both}u\mathrm{and}\mathrm{m},\mathrm{it}\mathrm{also}\mathrm{divides}v,$$$$\mathrm{hence}\mathrm{gcf}(u,m)\mathrm{divides}\mathrm{gcf}(v,m)$$$$\Rightarrow \mathrm{gcf}(u,m)\mid \mathrm{gcf}(v,m)$$$$$$$$\mathrm{We}\mathrm{know}\mathrm{if}a\mid b\mathrm{and}b\mid a,\mathrm{then}a=b\mathrm{or}a=-b.$$$$\mathrm{In}\mathrm{our}\mathrm{case},\mathrm{obviously}\mathrm{gcf}(u,m)=\mathrm{gcf}(v,m),\mathrm{since}$$$$\mathrm{gcf}(p,q){\mathrm{can}}^{\prime }\mathrm{t}\mathrm{be}\mathrm{negative}$$

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