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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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$k$ can be negative , so my solution is wrong.

I just realized thatLog in to reply

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$k$ 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 $t$.

I have found solution whenPS 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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