Let 𝑎, b, c ∈ R ∖ {0} satisfy the system:

𝑎 + 1/b = b + 1/c
b + 1/c = c + 1/𝑎

Find all possible values of 𝑎 + b + c. No graphing calculator. No WolframAlpha. Just raw algebra. Prove you're not a brainlet.

is this a joke? did you came up with this youself

>>16700645 (OP)
funny how these type of math olympiad questions become childishly trivial by calculus keeeeeek
chinese boys study 17 hours a day just to earn gold medals in IMO by solving these type of questions. Misérables

>>16700645 (OP)
the only answer is 3. proof: it came to me in a dream.

>>16700645 (OP)
you can manipulate both equations to get

[math]
a + 1/b - b = 1/c\\

b - 1/a = c - 1/c
[/math]

next f looks kinda invertible by the calculus, where f and g are [math]R^2[/math] valued functions

[math]
f(a,b) = g(c)\\

\Rightarrow (a,b) = (f^{-1} \circ g)(c) = h(c)
[/math]

so c is determined by arbitrary a and b, so the sum is actually a function h of c which you can find sort of idk senpai you do it

>>16700645 (OP)
You must be 18 to post here. Ask ChatGPT to solve your homework, pal

>>16701009
you can show the equations are cyclic, implying a = b = c are the only unique solutions, which suggests that all integer multiples of 3 are the general (and only) solutions for a + b + c.

>>16701017
my first thought was that "cyclical nature" but I've never used something like that to prove anything, I just remember a physics professor saying that can be exploited in 3-dim vector calculus

>test problem by shoving in 1s
>yep, it's retarded
Welp.

>>16701017
who said they were integers, any reals where a=b=c except 0 work. The next issue is, is there any soultion where a+b+c=0

>>16700685
>did you came up with this youself
ESL

>>16700645 (OP)
>Find all possible values of a + b + c
Let r != 0 and set a=b=c=r/3. Then the equations are satisfied and a+b+c=r.

I think you copied the problem wrong.

>>16702850
x, 1/(1-x), 1-1/x where x is the real root of x^3-3x+1=0.