Play with polynomial

The answer is yes. I proved this by searching for all solutions of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c}$ and checking whether they are also solutions of $\frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n} = \frac{1}{a^n+b^n+c^n}$

Step 1: Remove the fractions in the equation.

Fractions are not so nice. So, my first step is to multiply the equation by $abc(a+b+c)$ . Then we get $(a+b+c)(ab+ac+bc) = abc$ .

Intermezzo: How I found the solution..

This is not part of the proof, but it is always nice to show how you find the answer. In this case I simply wanted to find some small solutions. So, I tried $a = 1$ and $b=2$ and looked which $c$ would work. In this case I got the equation $(3 + c)(2 + 3c) =2c$ . This gave me the quadratic equations $3c^2+9c+6=0$ , which simplifies to (c+2)(c+1)=0. So, the solutions are c = -1 (= -a) or c = -2 ( =-b). After that, I also tried $a=2$ and $b=3$ which gave me a similar quadratic equation which simplifies to (c+2)(c+3)=0. So, my conjecture was that I would get (c+a)(c+b) = 0 for general a and b. That is what I started to prove.

Step 2: Solve $(a+b+c)(ab+ac+bc) = abc$ for $c$ .

I remove the brackets: $a^2b + a^2c + b^2a+ b^2c + c^2a + c^2b + 3abc = abc$ . I can rewrite this equation as a quadratic equation in the variable $c$ . That is $(a+b)c^2 + (a^2 + 2ab + b^2) c + (a^2b + b^2a) = 0$ . When $a \neq -b$ , we can divide this equation by $(a + b)$ . This gives us $c^2 + (a+b)c + ab$ . So, $(c+a)(c+b) = 0$ . So, the solutions are of the form $c = -a$ or $c = -b$ .

We have now shown that all solutions are of the form (a, b, c) = (x, -x, y), (y, x, -x) or (x, y , -x).

Step 3: Checking whether (a, b, c) = (x, -x, y) satisfies $\frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n} = \frac{1}{a^n+b^n+c^n}$

Filling in (a, b, c) = (x, -x, y) gives $\frac{1}{x^n} + \frac{1}{(-x)^n} + \frac{1}{y^n} \stackrel{?}{=} \frac{1}{x^n+(-x)^n+y^n}$ . For odd $n$ , we have $(-x)^n = -x^n$ . Due to this the equation simplifies to $\frac{1}{x^n} - \frac{1}{(x)^n} + \frac{1}{y^n} \stackrel{?}{=} \frac{1}{y^n}$ . Which is true for all nonzero $x, y$ .

Let $f(x)=x^3+mx^2+nx+p$ be a polynomial which has the roots $a,b,c$ . The given equality can be written as $(a+b+c)(ab+bc+ac)=abc$ .

So, we get $p=mn$ . Thus $f(x)=x^3+mx^2+nx+mn=(x+m)(x^2+n)$ . We see that one of the roots, say $a$ , is $-m=a+b+c,$ hence $c=-b$ . In that case $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{(-b)^n}=\frac{1}{a^n+b^n+(-b)^n}$ holds true for all odd integers $n$ .

Re-arrange the equation to get $\frac{1}{x+y+z} - \frac{1}{x} - \frac{1}{y} - \frac{1}{y} = 0$ . The algebra is tedious but the left hand side can be written in a productive way, yielding $\frac{-(x+y)(x+z)(y+z)}{xyz(x+y+z)} = 0$ . For this equation to be true, the numerator has to be 0 and this is only the case if $x=-y$ or $y = -z$ or $z = -x$ . And remember that none of them can 0. Without loss of generality let's suppose that $x=-y$ and $z$ is some non-zero real number. Then $\frac{1}{x^n} + \frac{1}{y^n} + \frac{1}{z^n} = \frac{1}{x^n} - \frac{1}{x^n} + \frac{1}{z^n} = \frac{1}{z^n} = \frac{1}{x^n - x^n + z^n} = \frac{1}{x^n + y^n + z^n}$ . As long as $n$ is odd, obviously.

See, $\frac {1}{a}+\frac {1}{b}+\frac {1}{c}=\frac {1}{a+b+c}$
$\Rightarrow (a+b+c)(ab+ac+bc)=abc$
$\Rightarrow (a+b)(b+c)(c+a)=0$
We have now shown that all solutions are of the form (a, b, c) = (x, -x, y), (y, x, -x) or (x, y , -x). It can be easily checked that this satisfies the given equation, for all or odd n, as checked by others.

Multiply across by $a$ $1+\frac{a}{b}+\frac{a}{c} = \frac{a}{a+b+c} = \frac{1}{1+\frac{b}{a}+\frac{c}{a}}$

Let $x=\frac{b}{a}$ and $y=\frac{b}{a}$ for simpler algebra. Multiply out: $(1+\frac{1}{x}+\frac{1}{y})(1+x+y)=1 \\ (xy+y+x)(1+x+y)=xy \\ 3xy+x+y+x^2+y^2+x^2y+xy^2=xy \\ x^2(y+1)+x(y^2+2y+1)+(y^2+y)=0 \\ x^2(y+1)+x(y+1)^2+y(y+1)=0 \\ (y+1)(x^2+x(y+1)+y)=0 \\ (y+1)(x+1)(x+y)=0$

So we get

$y=-1\Rightarrow c=-a$ , or
$x=-1\Rightarrow b=-a$ , or
$x=-y\Rightarrow b=-c$ .

Assume WLOG $b=-a$ . Then for odd $n$ , $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n}-\frac{1}{a^n}+\frac{1}{c^n}=\frac{1}{c^n}$ while, $\frac{1}{a^n+b^n+c^n}=\frac{1}{a^n-a^n+c^n}=\frac{1}{c^n}$ .

This holds regardless of the choice of $a,b,c$ .

The condition we rewrite in the following way: $c^2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+c\left(\dfrac{a}{b}+\dfrac{b}{a}+2\right)+a+b=0.$ Solving this equation in $c$ , we obtain $c=-a$ or $c=-b$ . Thus, for any positive odd integer $n$ , there holds $c^n=-a^n$ or $c^n=-b^n$ . So, taking, for instance, $c=-a$ , we have $\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{-a^n}=\dfrac{1}{b^n}=\dfrac1{a^n+b^n-a^n}=\dfrac1{a^n+b^n+c^n}$ .

Clearing denominators, we find that $(a+b)(b+c)(c+a) = 0.$ Clearing denominators of the second equation, we get $(a^n + b^n)(b^n + c^n)(c^n + a^n) = 0.$ But this equation is true because since $n$ is odd, we can factor out $(a+b)(b+c)(c+a)$ .

Answer: Yes.

Assume: $\frac{1}{a}$ + $\frac{1}{b}$ + $\frac{1}{c}$ = $\frac{1}{a+b+c}$

Multiplying by the denominators gives $(bc+ca+ab)(a+b+c) = abc$

Expanding gives $3abc+a^2(b+c)+b^2(c+a)+c^2(a+b) = abc$

One can see that $(a+b)(b+c)(c+a)=(a+b)(ab+ac+bc+c^2)=a^2(b+c)+b^2(c+a)+c^2(a+b)+2abc$

Hence $(a+b)(b+c)(c+a)=0.$

Replacing $a$ with $a^n$ etc. we can see that $\frac{1}{a^n}$ + $\frac{1}{b^n}$ + $\frac{1}{c^n}$ = $\frac{1}{a^n+b^n+c^n}$ if and only if $(a^n+b^n)(b^n+c^n)(c^n+a^n)=0.$

With $n$ odd, $(a+b)$ is a factor of $(a^n+b^n)$ , so $(a+b)(b+c)(c+a)$ is a factor of $(a^n+b^n)(b^n+c^n)(c^n+a^n)$ .

$$

Therefore $\frac{1}{a}$ + $\frac{1}{b}$ + $\frac{1}{c}$ = $\frac{1}{a+b+c}$ implies $(a^n+b^n)(b^n+c^n)(c^n+a^n)=0$ (for $n$ odd)

Thus implying $\frac{1}{a^n}$ + $\frac{1}{b^n}$ + $\frac{1}{c^n}$ = $\frac{1}{a^n+b^n+c^n}$ .

to make 1/a+1/b+1/c=1/(a+b+c) possible a,b,c one is negative (for e.g. a) one is same as negative number but x(-1) (for e.g. b) the remaining one can be anything(for e.g. c)

so no matter what power u add to them, if the power is the same, a^n+ b^n =0 so answer will always be c^n

Very easy, of course it can