An algebra problem by Rajdeep Ghosh
Suppose that a , b , c are non-zero complex numbers that satisfy
a 2 = b + c b 2 = c + a c 2 = a + b
Find the value of a + 1 a + b + 1 b + c + 1 c .
The answer is 2.
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1 solution
For completeness, the first equation should add " = (b+c) / (b+c+a) ".
Are there real (or complex) numbers that satisfy the conditions in the question? Otherwise, the question is vacuous.
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I expected people to solve it from there itself. Yes they are all real nos.
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Chung raises a good point. Can you check the reports to your question? They deal with the issue of existence of solutions under various conditions.
Note: You will require a , b , c = 0 in order to multiply the numerator and denominator by a , b , c in your equations.
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@Calvin Lin – I had immediately given the condition that a,b and c are positive distinct real nos. after @Chung Kevin had questioned about it. If Brilliant staff wants more changes, they can do it likewise.
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@Rajdeep Ghosh – The reason I commented, is that you didn't change it to "distinct real numbers". Instead, you changed it to "distinct natural numbers", which has no solution to the given equations, hence the reports.
As such, please clarify which version you intend.
Also, please update your solution to indicate why we are allowed to multiply the numerator and denominator by a a . In particular, we have to show that a b c = 0 , which is in a report.
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@Calvin Lin – If I am not wrong a/a(abc is not equal to zero anyway as a,b,c are distinct positive reals) is 1 so multiplying by 1 doesn't change the math altogether. I will change the condition right away.
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@Rajdeep Ghosh – Note that you only said "distinct real numbers" and not "distinct positive reals".
If you want to restrict to "distinct positive reals", you still have to demonstrate that there is a set of solutions.
In particular, I believe that there are only 8 solutions to the system, of which 1 is a = b = c = 0 , 1 is a = b = c = 2 , and the other 6 are cyclic permutations of { − 1 − i , i , i } .
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@Calvin Lin – But the question states that the nos. are distinct so there is no chance of them being equal. And I would also like to ask that do the other 6 values of a,b and c change the question altogether at last.
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@Rajdeep Ghosh – The point that I'm trying to make is that there is no system of "distinct positive reals" that satisfy the conditions in your question. As such, the final question is moot.
Instead of using "distinct positive reals", all that you need is " a b c = 0 ". That would allow you to immediately justify why we can multiply by a a without introducing any errors. This way, the solution that you found holds true for all possible solution sets that satisfy a b c = 0 , even the 6 complex ones that I stated.
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@Calvin Lin – I've edited the problem for clarity. You should see how I've phrased it, to avoid these issues that I've rasied above.
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@Calvin Lin – Thank You Very Much. I was just trying to find a god way to frame the question. I will keep the points in mind.
a/(a+1)=a^2/(a^2+a)=(b+c)/(a+b+c) Similarly for b and c. So, (b+c)/(a+b+c)+(a+c)/(a+b+c)+(a+b)/(a+b+c)=2