For (a,b,c) belong to \(\mathbb{Z}\), I have a solution..
I think the solution for (a,b,c) is of the form:

(-4-t, t, 2) for some integer t . My solution:
\[ac+bc+c^2-a-b=0\]
\[\Rightarrow (ac-a)+(bc-b)+(c^2-1)=-1\]
\[\Rightarrow (c-1)(a+b+c+1)=-1\]
For product of two integers to be -1 they must be of the form (-1,1) or vice versa....
In our case c-1=-1 can be easily rejected since for that value RHS is turning to undefined.. Hence
c-1=1 and a+b+c+1=-1 and hence the ordered triplet for (a,b,c) is (-4-t,t ,2) for some integer t.

@Nihar Mahajan
–
Yeah... You had to take k=-2 for c to become 2..... Anyways Is my solution complete? ( I'm just a beginner in solving Diophantine equation :-})..Might be I have missed some cases too!!

@Rishabh Jain
–
I have found solution when \(k\) is negative. Let \(k=-m\) for some positive integer \(m\) , then \(c=\dfrac{m}{m-1}\). Again let \(m=n(m-1)\) for some integer \(n\) , so we have \(\dfrac{1}{m}+\dfrac{1}{n}=1\) and the only integer solution for this is \(m=n=2\). Substituting it we have \(c=2\) , \(a+b=-4\) so the solution set is \((a,b,c)=(-4-t,t,2)\) for some integer \(t\).

PS your solution is shorter , because you are cool :P

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## Comments

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TopNewestFor (a,b,c) belong to \(\mathbb{Z}\), I have a solution..

I think the solution for (a,b,c) is of the form:

(-4-t, t, 2) for some integer t . My solution:

\[ac+bc+c^2-a-b=0\] \[\Rightarrow (ac-a)+(bc-b)+(c^2-1)=-1\] \[\Rightarrow (c-1)(a+b+c+1)=-1\] For product of two integers to be -1 they must be of the form (-1,1) or vice versa....

In our case c-1=-1 can be easily rejected since for that value RHS is turning to undefined.. Hence

c-1=1 and a+b+c+1=-1 and hence the ordered triplet for (a,b,c) is (-4-t,t ,2) for some integer t.

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You may use the fact that \(c\) divides \(a+b\) .

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Is that helpful in finding the solution?

Because I seem to have tried it before and it didn't work out for me.

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Let \(a+b=ck\) for some integer \(k\). Then \(ck+c=k \Rightarrow c=\dfrac{k}{k+1}\) .

<The remaining solution is completed below>

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PS your solution is shorter , because you are cool :P

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(You should have declared if a,b,c are integers or just real) If they are real then a=b=c=2/3 is a solution

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Oh yeah. That is a possible solution.

What would it be for integers?

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