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# A little twist to an old problem

$\large a+b+c= \dfrac {a+b}{c}$

Find all the solutions to the diophantine equation above.

Note by Mehul Arora
9 months, 3 weeks ago

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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. · 9 months, 3 weeks ago

You may use the fact that $$c$$ divides $$a+b$$ . · 9 months, 3 weeks ago

Is that helpful in finding the solution?

Because I seem to have tried it before and it didn't work out for me. · 9 months, 3 weeks ago

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

<The remaining solution is completed below> · 9 months, 3 weeks ago

Haven't you missed a=-4-t, b=t, c=2 (for some integer t) ... Hence the equation would have infinite integral solutions · 9 months, 3 weeks ago

I just realized that $$k$$ can be negative , so my solution is wrong. · 9 months, 3 weeks ago

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!! · 9 months, 3 weeks ago

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 · 9 months, 3 weeks ago

(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 · 9 months, 3 weeks ago

Oh yeah. That is a possible solution.

What would it be for integers? · 9 months, 3 weeks ago