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Let $m$ and $n$ be relatively prime integers different from $\pm 1$. Show that there are unique integers $u$ , $v$ $\in Z$ such that $um+vn=1$ and $0 \le u \lt |n|$. In this case show that $|v| \lt |m|$

My try: Let there be two such integers, $u$, $u'$, $v$ and $v'$. Then $um+vn=1$ and $u'm+v'n=1$. Subtracting one from the other we get $m(u-u')+n(v-v')=0$. Now $m(u-u')=n(v'-v)$. Since $(m,n)=1$, we have $n | (u-u')$. Also $m |(v-v')$. There exists $k$ and $k'$ such that $u=u'+nk$ and $v=v'+mk'$.

From here how do I show the required??

Thanks for the help!!

hardmath
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tattwamasi amrutam
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    We need to show two things here, existence and uniqueness. The existence part does not seem to have been addressed. I do not know what can be assumed here, since the problem presumably arises in a certain course context. I would guess that it has already been proved that there exist integers $x$ and $y$ such that $xm+yn=1$, and we are only asked to deal with size issues. – André Nicolas Aug 06 '14 at 13:19
  • @AndréNicolas how do i show existence?? – tattwamasi amrutam Aug 08 '14 at 09:06
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    I will assume you know there are $x_0,y_0$ such that $x_0m+ny_0=1$. Then for any $t$, we have $(x_0-tn)m+(y_0+tm)n=1$. By the Division Algorithm, there is a $t$ such that $x_0-tn$ satisfies the desired inequalities. – André Nicolas Aug 08 '14 at 13:26
  • @AndréNicolas got it – tattwamasi amrutam Aug 09 '14 at 01:41
  • Good! The uniqueness has been amply dealt with in the answers. – André Nicolas Aug 09 '14 at 01:45
  • @AndréNicolas Now the new $x_0=x_0-tn$. But I also need to show that $|y_0+tm|\lt |m|$ simultaneously. How do I show it?? – tattwamasi amrutam Aug 09 '14 at 02:11
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    We don't need to use $t$ for the second part. If $0\lt u\lt |n|$, then $u|m|\lt |mn|$, so $u|m|-1\lt |mn|-1$. But $u|m|-1=-v|n|$, so $|vn|\lt |mn|-1$, which forces $|v|\lt |m|$. – André Nicolas Aug 09 '14 at 02:29

2 Answers2

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If $u,u'$ both satisfy $0 \le x < |n|$ then the absolute value of their difference is less than $|n|$. You have already arrived at $u=u'+nk$ so the absolute value of their difference is $|n||k|.$ So now what if $k\neq 0$?

coffeemath
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Hint $\ $ You've proved $\,u'\equiv u\pmod n.\,$ Therefore their remainders mod $\,n\,$ are equal. However, they equal their remainders mod $\,n\,$ since $\,0\le u',u < |n|.\,$ Therefore they are equal $\,u' = u.$

Remark $\ $ Such consequences of the uniqueness of the remainder in the Division Algorithm are ubiquitous in number theory and algebra, so it's very worthwhile to be familiar with this viewpoint. For example here is one of my favorites.

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Bill Dubuque
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