Let $a,b \in \Bbb N $ with $\gcd(a,b)=1$. The equation $ax + by = ab$ has the obvious solution $(b, 0)$ in integers. Show, however, that it has no solution in positive integers.
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What happens when looking modulo $b$? And $a$? – Lord_Farin Oct 18 '12 at 14:39
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1Does this mean $by=moda$ and $ax=modb$? – blah Oct 18 '12 at 14:46
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Yes, $by = 0 \pmod a$ and $ax = 0 \pmod b$. Now you can use what you know about $\gcd(a,b)$. – Lord_Farin Oct 18 '12 at 14:49
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So because $\gcd(a,b)=1$, $y$ must divide $a$ and $x$ must divide $b$? – blah Oct 18 '12 at 14:52
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Hint $\ $ By Euclid's Lemma, $\rm\ (a,b) = 1,\,\ a,b\:|\:ax+by\:\Rightarrow\: a\:|\:y,\ b\:|\:x\:\Rightarrow\:ab\:|\:ax,by$
Bill Dubuque
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so $ax=kab$ for some $k \in \Bbb Z$ and $by=nab$ for some $n \in \Bbb Z$ which means $k+n=1$ and aside from the given case this means that either $k$ or $n$ is negative? Is this correct? – blah Oct 18 '12 at 15:15
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Yes, that's one way to conclude. Or: if two common multiples $\ge 0$ have sum = lcm then one of them is $0$ and the other is the lcm, by the leastness of the lcm. – Bill Dubuque Oct 18 '12 at 15:18
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@jjagmath Actually it is - I'm using the general divisibility form of Euclid's Lemma (see the link I added). Please wait for clarification before downvoting. – Bill Dubuque May 23 '25 at 14:01
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Your reference is another post from you? So you agree with yourself? What about a reference from a book? – jjagmath May 23 '25 at 14:11
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@jjagmath If you prefer to restrict the name "Euclid's Lemma" to the case of prime divisor then that is certainly your prerogative. Some authors do use it in this more general sense. I've been using it heavily here for over $15$ years and you are the first to have any issue with it. I'm a bit shocked that you would downvote due to terminology (and before waiting for clarification). $\ \ $ – Bill Dubuque May 23 '25 at 14:19
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Assume x and y are integers. $ax + by =ab$ is equivalent to $a(x-b)=b(a-y)$. Since gcd(a,-b)=1, a must divide into a-y. Thus, $a-y=ak$, or $y=a(1-k)$ where k is an integer (positive or negative). Now we can write $a(x-b)=b(a-y)=b(ak)$. Canelling out the a's on each side, we now have $x-b=bk$. So $x=b(1+k)$. Thus an integer solution must be of the form $(b(1+k);a(1-k))$ with k being an integer. We want both x and y to be positive. x is positive if and only if $k \ge -1$ and y is positive if and only if $1 \ge k$. Thus, k can only be equal to -1, 0 or 1. If k=0, we have (b,0) as a solution, if k=1, the solution is (0,a). Other than those two solutions, there are no positive integer solutions. k=-1 makes y negative.
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The general integer solution is $x=b+tb$, $y=-ta$. We're assuming $a,b,x,y>0$, so we have from the $x$ equation that $1+t>0$ and from the $y$ equation that $-t>0$. Putting these together gives $-1<t<0$, but there are no integers $t$ in that range.
coffeemath
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Assume $a\geq 2$. Since $ax-ab=-by$ therefore $a|by$. But $gcd(a,b)=1$ thus $a|y$. Hence $y=0,a,2a,\ldots$. The only possibilities $y=0,a$. If $y=0$ then $x=b$, if $y=a$ then $x=0$. If $a=1$ then by symmetry if $b\geq 2$ we obtain the statement. The case $a,b=1$ is trivial.
vesszabo
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