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Consider the Diophantine equation $ \ px+qy=n \ $ , where $ \ p , q \in \mathbb{Z}^+ $ and

$ \ n \in \mathbb{Z}^- $.

Let $ \ \gcd(p,q) \ $ be a factor of $ \ n \ $, i.e. $ \ \gcd(p,q) \ =\ nk $ with $k\in\mathbb{Z}$

Show that if $ \ n >pq-(p+q) \ $ then $ \ px+qy=n \ $ has at least one solution in positive integers .

Answer:

Let $ \ \gcd(p,q)=d \ $

Then $ \ d \ |n \ $ and which implies $ \ n=kd \ $ for some $ \ k \in \mathbb{Z} \ $.

Further since $ \ \gcd(p,q)=d \ $ , there exists $ \ x' , y' \in \mathbb{Z} \ $ such that

$$ \ px'+qy'=d. \\ \text{Multiplying by $k$ gives} , \\ k(px'+qy')=kd \\ \Rightarrow p(kx')+q(ky')=n \ $$

Let $ \ x_0=kx', \ y_0=ky' \ $ , then

$ px_0+qy_0=n \ $ giving solution $ \ (x_0,y_0) \ $

But how to use the given condition in order to get positive integer solutions i.e. $ \ x_0 , \ y_0 \ $ will be positive integer ?

Help me out

Algebear
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MAS
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1 Answers1

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You should look up the Euclidean algorithm: https://en.wikipedia.org/wiki/Euclidean_algorithm#Worked_example You can use this to determine the values in the formula of Bézout, that is $\gcd(p,q)=px+qy=:d$ with $x,y\in\mathbb{Z}$.

Algebear
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  • would you do little more. Because I am not clear at this – MAS Apr 26 '18 at 17:15
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    Hmm, it just is a lot of work. I could do it, but you learn way more if you apply the Euclidean algorithm. Searching on google for this algorithm will give plenty of explanations/tutorials for applying it. I went through the same once:-)) – Algebear Apr 26 '18 at 17:26
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    Maybe you can use that $pq−(p+q)=(1-p)(1-q)−1 (I'm not sure, but writing things shorter usually helps). Sorry, previous comment was non-sense from my side...(which I now deleted) – Algebear Apr 26 '18 at 17:37