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If $a,b,c$ are three distinct real numbers and

$$a+\frac1b=b+\frac1c=c+\frac1a=t$$

for some real number $t$ prove that $abc+t=0$

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We can use $c = t - 1/a$ to eliminate $c$ from the set of three equations. We obtain: $$ (t - b) (t - 1/a) = 1 \\ a = t - 1/b $$ Using the second formula to eliminate $a$ from the first yields: $$ t^3 - t^2 (b + 1/b) - t + (b + 1/b) = 0 $$ This third order equation in $t$ can be rewritten as follows. $$ (t + 1) (t - 1) (t - b - 1/b) = 0 $$ We see that $t$ has three solutions: $t = 1$, $t = -1$ and $t = b + 1/b.$

(I) $t = 1$. This leads to the solution: $a = x$, $b = 1/(1-x)$, $c = (x-1)/x$ with $x$ a real number in $(-\infty, +\infty)$. The product $abc$ equals $-1$, hence the solution is in agreement with $abc + t = 0$.

(II) $t = -1$. This leads to the solution: $a = x$, $b = -1/(1+x)$, $c = -(1+x)/x$. Again $x$ is a real number in $(-\infty, +\infty)$. The product $abc$ equals $+1$. Since $t = -1$, in the solution is in agreement with $abc + t = 0$.

(III) $t = b + 1/b$. This leads to the solution: $a = x$, $b = x$, $c = x$, with $x$ a real number in $(-\infty, +\infty)$. The product $abc$ equals $x^3$. Since $t = x + 1/x$, this solution is not in agreement with $abc + t = 0$. However, the problem states that $a$, $b$ and $c$ must be distinct. On that ground we are forced to omit this solution.

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  • $\begingroup$ I reformatted your answer yo make it easier to read. I also corrected an error in part (II). Feel free to undo my edits if they seem unjust. $\endgroup$ – Joonas Ilmavirta Nov 22 '14 at 11:59
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Based upon the symmetry of the equalities, I would guess that $a$, $b$, $c$ are identical values. Let's see if that's right - I have no mathematical evidence to back that up at this point.

To check my guess, I will do a simple substitution. $$\frac{ab+1}{b}=t, \frac{bc+1}{c}=t, \frac{ca+1}{a}=t$$ Solving the original equalities for the three variables of interest gives: $$a=t-\frac{1}{b}=\frac{bt-1}{b},b=t-\frac{1}{c}=\frac{ct-1}{c},c=t-\frac{1}{a}=\frac{at-1}{a}$$ Start doing the substitution into the second expression. $$\frac{bt-1}{b}*\frac{ct-1}{c}*\frac{at-1}{a}+t=0$$ Put over common denominator: $$(bt-1)(ct-1)(at-1)+abc*t=0$$ Expand: $$abc*t^3-ab*t^2-ac*t^2-bc*t^2+at+bt+ct-1+abc*t=0$$ At this point, we have a cubic equation. I am not certain if there is a trivial factorization of this completely, but we don't need that. By the fundamental theorem of algebra, there exists at least one real-valued $t$ for which the above equation holds.

I am going to see if I can figure out what it is. Here we go. $$abc*t^3+(-ab-ac-bc)*t^2+(a+b+c+abc)*t-1=0$$ Wolfram Alpha solution is this: $$t = (-9 a^3 b^3 c^2+2 a^3 b^3-9 a^3 b^2 c^3-3 a^3 b^2 c-3 a^3 b c^2+2 a^3 c^3-9 a^2 b^3 c^3-3 a^2 b^3 c+12 a^2 b^2 c^2-3 a^2 b c^3+\sqrt((-9 a^3 b^3 c^2+2 a^3 b^3-9 a^3 b^2 c^3-3 a^3 b^2 c-3 a^3 b c^2+2 a^3 c^3-9 a^2 b^3 c^3-3 a^2 b^3 c+12 a^2 b^2 c^2-3 a^2 b c^3-3 a b^3 c^2-3 a b^2 c^3+2 b^3 c^3)^2+4 (3 a b c (a b c+a+b+c)-(-a b-a c-b c)^2)^3)-3 a b^3 c^2-3 a b^2 c^3+2 b^3 c^3)^(1/3)/(3 2^(1/3) a b c)-(2^(1/3) (3 a b c (a b c+a+b+c)-(-a b-a c-b c)^2))/(3 a b c (-9 a^3 b^3 c^2+2 a^3 b^3-9 a^3 b^2 c^3-3 a^3 b^2 c-3 a^3 b c^2+2 a^3 c^3-9 a^2 b^3 c^3-3 a^2 b^3 c+12 a^2 b^2 c^2-3 a^2 b c^3+\sqrt((-9 a^3 b^3 c^2+2 a^3 b^3-9 a^3 b^2 c^3-3 a^3 b^2 c-3 a^3 b c^2+2 a^3 c^3-9 a^2 b^3 c^3-3 a^2 b^3 c+12 a^2 b^2 c^2-3 a^2 b c^3-3 a b^3 c^2-3 a b^2 c^3+2 b^3 c^3)^2+4 (3 a b c (a b c+a+b+c)-(-a b-a c-b c)^2)^3)-3 a b^3 c^2-3 a b^2 c^3+2 b^3 c^3)^(1/3))-(-a b-a c-b c)/(3 a b c)$$

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  • $\begingroup$ how could you say that there is one real valued 't' for which the cubic equation holds $\endgroup$ – lokesh sangabattula Nov 16 '14 at 11:33
  • $\begingroup$ a,b,c are real valued , the for any root of the above equation its complex conjugate is also a root. Thus equation roots occur in conjugate pairs. The only way in which odd number of roots is possible is if odd number of the roots were real. Thus at least one root is real $\endgroup$ – Abhishek Shetty Nov 16 '14 at 11:47
  • $\begingroup$ ok i understood $\endgroup$ – lokesh sangabattula Nov 16 '14 at 15:26
  • $\begingroup$ from the original question: "a,b,c are three DISTINCT real numbers". $\endgroup$ – Will Sherwood Nov 16 '14 at 21:41
  • $\begingroup$ Woops, good catch, @WillSherwood, I don't know what I was thinking when I wrote that originally. Haha. $\endgroup$ – FundThmCalculus Nov 17 '14 at 1:42

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