\[ \large a+b+c= \dfrac {a+b}{c} \]

Find all the solutions to the diophantine equation above.

\[ \large a+b+c= \dfrac {a+b}{c} \]

Find all the solutions to the diophantine equation above.

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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. – Rishabh Cool · 1 year, 2 months ago

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You may use the fact that \(c\) divides \(a+b\) . – Nihar Mahajan · 1 year, 2 months ago

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Because I seem to have tried it before and it didn't work out for me. – Mehul Arora · 1 year, 2 months ago

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<The remaining solution is completed below> – Nihar Mahajan · 1 year, 2 months ago

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– Rishabh Cool · 1 year, 2 months ago

Haven't you missed a=-4-t, b=t, c=2 (for some integer t) ... Hence the equation would have infinite integral solutionsLog in to reply

– Nihar Mahajan · 1 year, 2 months ago

I just realized that \(k\) can be negative , so my solution is wrong.Log in to reply

– Rishabh Cool · 1 year, 2 months 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!!Log in to reply

PS your solution is shorter , because you are cool :P – Nihar Mahajan · 1 year, 2 months 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 · 1 year, 2 months ago

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What would it be for integers? – Mehul Arora · 1 year, 2 months ago

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