a, b, c are distinct reals such that $$a + 1/b = b + 1/c = c + 1/a = t $$for some real t. Show that t = -abc I tried using continued fractions to isolate a,b and c but equations of degree more than 2 are formed.
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This is just quite laborious. We firstly randomly select a variable as free variable, say this time we choose $a$. Treat the condition as three separate ones. We have: $$a+1/b=t \text{ in which we solve for b}$$ $$c+1/a=t \text{ in which we solve for c}$$ $$b+1/c=t \text{ in which we solve for c with b obtained from (1)}$$ And the $c$ obtained from the two equation should be the same. We got a new equation with only $a$ and $t$: $$\frac {a-t}{ta-{t}^{2}+1}=\frac {ta-1}{a}$$ which is equivalent to: $$({t}^{2}-1){a}^{2}+(t-{t}^{3})a+{t}^{2}-1=0$$ Solving this yields t can equal to any of $1,-1,\frac {{a}^{2}+1}{a}$.
However, they are said to be three distinct reals (or $a=c$), so $t=1,-1$. (-1 seems to be incorrect in the next step, not sure about my calculation)
Now you substitute the possible values of $t$ to the original equation. Solve the three equations and verify the result.
It seems $t$ is just a trap that since it is fixed by the three conditions as implied in the comment.
YiFei
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Why can't $t=\frac{a^2+1}{a}$ – 420 Sep 07 '16 at 08:11
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If $t=\frac{a^2+1}{a}$, then $t=a+1/a=c+1/a$, contradicts with your condition that they are distinct reals. – YiFei Sep 07 '16 at 08:13
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Did you actually solve the cubic in t.If yes then how? – 420 Sep 07 '16 at 08:16
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@user348631 I didn't actually solved the equation, I did a long division over t-1. – YiFei Sep 07 '16 at 08:20
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What did you divide by$ t-1$ – 420 Sep 07 '16 at 08:27
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Just see my updated answer, which include a equation in which you can clearly see you can factor out $t^2-1$ actually. – YiFei Sep 07 '16 at 08:29