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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
7 months, 2 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. Rishabh Cool · 7 months, 2 weeks ago

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You may use the fact that \(c\) divides \(a+b\) . Nihar Mahajan · 7 months, 2 weeks ago

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@Nihar Mahajan Is that helpful in finding the solution?

Because I seem to have tried it before and it didn't work out for me. Mehul Arora · 7 months, 2 weeks ago

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

<The remaining solution is completed below> Nihar Mahajan · 7 months, 2 weeks ago

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@Nihar Mahajan Haven't you missed a=-4-t, b=t, c=2 (for some integer t) ... Hence the equation would have infinite integral solutions Rishabh Cool · 7 months, 2 weeks ago

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@Rishabh Cool I just realized that \(k\) can be negative , so my solution is wrong. Nihar Mahajan · 7 months, 2 weeks ago

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@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 Cool · 7 months, 2 weeks ago

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@Rishabh Cool 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 Nihar Mahajan · 7 months, 2 weeks ago

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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 Tudor Darius Cardas · 7 months, 2 weeks ago

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@Tudor Darius Cardas Oh yeah. That is a possible solution.

What would it be for integers? Mehul Arora · 7 months, 2 weeks ago

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