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Assuming that the system of equations:

$$x=yz$$ $$u=vw$$ $$s=rt$$

has no integers solutions. How I can prove that the equation $$ax+bu+cs=ayz+bvw+crt$$ also has no integers solutions for any integers $a,b,c$

Safwane
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  • I would start by assuming a solution to the equation exists, then finding out what that means for the system of equations. What have you tried so far? – abiessu Jan 11 '21 at 17:21
  • How about plugging things in? –  Jan 11 '21 at 17:22
  • @abiessu: In fact I have no idea to start. – Safwane Jan 11 '21 at 17:22
  • Where did you encounter this problem? Are there any similar problems you have worked on recently? – abiessu Jan 11 '21 at 17:25
  • @abiessu: This is just an exam problem. – Safwane Jan 11 '21 at 17:26
  • This being an exam problem suggests that there was likely some homework prior that was relevant. Can you recall any such homework? – abiessu Jan 11 '21 at 17:28
  • @abiessu: I am sure there is no such thing. – Safwane Jan 11 '21 at 17:29
  • Don't cheat on exams... –  Jan 11 '21 at 17:31
  • @newQOpenGLWidget: Student: I think he looked up the answer – Safwane Jan 11 '21 at 17:32
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    I'm not sure the question makes sense. And integer solution for $a,b,c$ means $a,b,c$ can't be integers for a fixed $x,y,z,u,v,w,s,t,r$ But the fixed $x,y,z,u,v,w,s,t,r$ could be anything. When you say $x=yz$ and $u=vw$ and $s =rt$ has no integer solutions. Now integer solutions to what? Which are the fixed values and which are the variables to solve for? Without more information we can set $y,z,v,w,r,t$ to any integers we want $x=yz;u=vw;s=rt$ will be integers and we can set $a,b,c$ to any integers we want. ... So what exactly is the question. What is set and what to solve? – fleablood Jan 11 '21 at 17:33
  • According to WolframAlpha $\qquad a=b=c=0\qquad$ – poetasis Jan 18 '21 at 01:05

1 Answers1

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$$ax+bu+cs=ayz+bvw+crt\\ a(x-yz)+b(u-vw)+c(s-rt)=0$$

Plugging in any specific integers for $x,y,z,u,v,w,r,s,t$ yields specific integer values $x-yz,u-vw,s-rt$. It really doesn't matter what these values are, they just become coefficients for $ae+bf+cg=0$ which is always solvable in integers (e.g., $a=b=c=0$). So the implication in the question is false as a starting point (i.e., solutions need not exist for the original "system" in order for the equation to have solutions).

After that, note that the original "system" is somewhat meaningless as there are no constraints given on the listed variables aside from "this system doesn't have solutions".

abiessu
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