If a,b,c are three distinct real numbers

Question

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

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

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

Answer 1

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.

Answer 2

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: $$t = (-9 a^3 b^3 c^2+2 a^3 b^3-9 a^3 b^2 c^3-3 a^3 b^2 c-3 a^3 b c^2+2 a^3 c^3-9 a^2 b^3 c^3-3 a^2 b^3 c+12 a^2 b^2 c^2-3 a^2 b c^3+\sqrt((-9 a^3 b^3 c^2+2 a^3 b^3-9 a^3 b^2 c^3-3 a^3 b^2 c-3 a^3 b c^2+2 a^3 c^3-9 a^2 b^3 c^3-3 a^2 b^3 c+12 a^2 b^2 c^2-3 a^2 b c^3-3 a b^3 c^2-3 a b^2 c^3+2 b^3 c^3)^2+4 (3 a b c (a b c+a+b+c)-(-a b-a c-b c)^2)^3)-3 a b^3 c^2-3 a b^2 c^3+2 b^3 c^3)^(1/3)/(3 2^(1/3) a b c)-(2^(1/3) (3 a b c (a b c+a+b+c)-(-a b-a c-b c)^2))/(3 a b c (-9 a^3 b^3 c^2+2 a^3 b^3-9 a^3 b^2 c^3-3 a^3 b^2 c-3 a^3 b c^2+2 a^3 c^3-9 a^2 b^3 c^3-3 a^2 b^3 c+12 a^2 b^2 c^2-3 a^2 b c^3+\sqrt((-9 a^3 b^3 c^2+2 a^3 b^3-9 a^3 b^2 c^3-3 a^3 b^2 c-3 a^3 b c^2+2 a^3 c^3-9 a^2 b^3 c^3-3 a^2 b^3 c+12 a^2 b^2 c^2-3 a^2 b c^3-3 a b^3 c^2-3 a b^2 c^3+2 b^3 c^3)^2+4 (3 a b c (a b c+a+b+c)-(-a b-a c-b c)^2)^3)-3 a b^3 c^2-3 a b^2 c^3+2 b^3 c^3)^(1/3))-(-a b-a c-b c)/(3 a b c)$$