A little twist to an old problem

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

Find all the solutions to the diophantine equation above.

Note by Mehul Arora
3 years, 8 months ago

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For (a,b,c) belong to Z\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+c2ab=0ac+bc+c^2-a-b=0 (aca)+(bcb)+(c21)=1\Rightarrow (ac-a)+(bc-b)+(c^2-1)=-1 (c1)(a+b+c+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 Jain - 3 years, 8 months ago

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You may use the fact that cc divides a+ba+b .

Nihar Mahajan - 3 years, 8 months ago

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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.

Mehul Arora - 3 years, 8 months ago

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

<The remaining solution is completed below>

Nihar Mahajan - 3 years, 8 months 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 Jain - 3 years, 8 months ago

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@Rishabh Jain I just realized that kk can be negative , so my solution is wrong.

Nihar Mahajan - 3 years, 8 months 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 Jain - 3 years, 8 months ago

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@Rishabh Jain I have found solution when kk is negative. Let k=mk=-m for some positive integer mm , then c=mm1c=\dfrac{m}{m-1}. Again let m=n(m1)m=n(m-1) for some integer nn , so we have 1m+1n=1\dfrac{1}{m}+\dfrac{1}{n}=1 and the only integer solution for this is m=n=2m=n=2. Substituting it we have c=2c=2 , a+b=4a+b=-4 so the solution set is (a,b,c)=(4t,t,2)(a,b,c)=(-4-t,t,2) for some integer tt.

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

Nihar Mahajan - 3 years, 8 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 - 3 years, 8 months ago

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Oh yeah. That is a possible solution.

What would it be for integers?

Mehul Arora - 3 years, 8 months ago

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