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Problem: Let $s$ and $g>0$ be given integers. Prove that integers $x$ and $y$ exist satisfying $x+y=s$ and $(x,y)=g$ iff $g|s$.

My Attempt I have already proved the theorem in the "right " direction. So I shall write the converse: Assuming $g|s$ then if we let $x=g$ and $y=s-g$, we get $(x,y)=(g,s-g)=g$ and $x+y=g+s-g=s.$ Thus there exists $x$ and $y$ such that the requirements of the problem are met.

Please tell me whether the proof of the converse is correct or not? Thanks in advance!

Student
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    Your argument looks good. – lulu Nov 02 '16 at 13:51
  • In similar spirit, if I am given the following problem: For $g$ and $l>0$ show that if $g|l$ then there exists integers $x$ and $y$ such that $(x,y)=g$ (GCD) and $[x,y]=l$ (LCM). Then can we prove this fact by claiming that if $x=g$ and $y=l$ we get $(x,y)=g$ and $[x,y]=l$. – Student Nov 02 '16 at 14:37

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