We have an answer to "If 1xb+x−c+1+1xc+x−a+1+1xa+x−b+1=1 then what can we say about..."

Malaki Aguilar

Answered question

2022-02-04

\frac{1}{x^{b} + x^{- c} + 1} + \frac{1}{x^{c} + x^{- a} + 1} + \frac{1}{x^{a} + x^{- b} + 1} = 1

then what can we say about a, b, c?

Alvaro Pollard

Beginner2022-02-05Added 16 answers

For any non-zero value of x, we have

x^{0} = 1

.
So with

a = b = c = 0

we have:

\frac{1}{x^{b} + x^{- c} + 1} + \frac{1}{x^{c} + x^{- a} + 1} + \frac{1}{x^{a} + x^{- b} + 1}

= \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1

Note also that if

a = b = c = k

then:

1 = \frac{1}{x^{k} + x^{- k} + 1} + \frac{1}{x^{k} + x^{- k} + 1} + \frac{1}{x^{k} + x^{- k} + 1}

= \frac{3}{x^{k} + x^{- k} + 1}

So we have:

x^{k} + x^{- k} + 1 = 3

Subtracting 3 from both sides and multiplying through by

x^{k}

we get:

0 = {(x^{k})}^{2} - 2 (x^{k}) + 1 = {(x^{k} - 1)}^{2}

.
So

x^{k} = 1

This is satisfied for any non-zero value of x if

k = 0

and no other values of k.

Sam Jensen

Beginner2022-02-06Added 13 answers

Using "brute force" or with the help of a symbolic processor,

\frac{1}{x^{b} + x^{- c} + 1} + \frac{1}{x^{c} + x^{- a} + 1} + \frac{1}{x^{a} + x^{- b} + 1} = \frac{n}{d} = 1

n = (x^{a} + x^{b} + 2 x^{a + b} + x^{2 a + b} + x^{c} + 2 x^{a + c} +

2 x^{b + c} + 6 x^{a + b + c} + 2 x^{2 a + b + c} + x^{2 b + c} +

2 x^{a + 2 b + c} + x^{2 a + 2 b + c} + x^{a + 2 c} +

2 x^{a + b + 2 c} + x^{2 a + b + 2 c} + x^{a + 2 b + 2 c})

d = (1 + x^{a} + x^{b} + 2 x^{a + b} + x^{2 a + b} + x^{c} + 2 x^{a + c} +

2 x^{b + c} + 4 x^{a + b + c} + 2 x^{2 a + b + c} + x^{2 b + c} +

2 x^{a + 2 b + c} + x^{2 a + 2 b + c} + x^{a + 2 c} +

2 x^{a + b + 2 c} + x^{2 a + b + 2 c} + x^{a + 2 b + 2 c} + x^{2 a + 2 b + 2 c})

but

n - d = - 1 + 2 x^{a + b + c} - x^{2} (a + b + c)

so if

n = d

- 1 + 2 x^{a + b + c} - x^{2 (a + b + c)} = 0

Solving for

x^{a + b + c}

we obtain

x^{a + b + c} = 1

then

a + b + c = 0

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