# Thread: If a then what is b

1. ## If a then what is b

Hi all,
I'm really stuck on this question.
At first I thought it was trivial but I have had second thoughts about that. Can someone help please.

If $\left({x+\frac{1}{x}+1}\right)\left({x+\frac{1}{x} }\right)=1$

then what is

$\left({x^{20}+\frac{1}{x^{20}}+1}\right)\left({x^{ 20}+\frac{1}{x^{20}}}\right)$

I know that you want to see what I have done first but I really haven't done anything that I see as helpful.
If the first statement was true for all real x the second one would be also equal to 1.
But this is not the case so I really don't know where to start.
Thankyou.

2. ## Re: If a then what is b

Originally Posted by Melody2
Hi all,
I'm really stuck on this question.
At first I thought it was trivial but I have had second thoughts about that. Can someone help please.

If $\left({x+\frac{1}{x}+1}\right)\left({x+\frac{1}{x} }\right)=1$

then what is

$\left({x^{20}+\frac{1}{x^{20}}+1}\right)\left({x^{ 20}+\frac{1}{x^{20}}}\right)$

I know that you want to see what I have done first but I really haven't done anything that I see as helpful.
If the first statement was true for all real x the second one would be also equal to 1.
But this is not the case so I really don't know where to start.
Thankyou.
symbols are symbols.

any restriction on x in your first equation? No? So subsitute x = $x^{20}$.

3. ## Re: If a then what is b

Multiplying out: $x^2+x+2+\dfrac{1}{x} + \dfrac{1}{x^2} = 1$.

Multiply both sides by $x^2$ to get $x^4 + x^3 + x^2 +x + 1 = 0$.

Multiplying both sides by $x-1$ gives $x^5-1 = 0$ (but $x \neq 1$ since we added the root). Hence, $x = -(-1)^{1/5}, x= (-1)^{2/5}, x = -(-1)^{3/5}, x = (-1)^{4/5}$ are the four solutions. In each case, $x^5 = 1$. So:

$\left(x^{20} + \dfrac{1}{x^{20}} + 1\right)\left(x^{20} + \dfrac{1}{x^{20}}\right) = \left((x^5)^4 + \dfrac{1}{(x^5)^4} + 1\right)\left( (x^5)^4 + \dfrac{1}{(x^5)^4}\right) = 3\cdot 2 = 6$

4. ## Re: If a then what is b

Thanks Romsek,
I am glad that your intitial thought was the same as mine. It makes me feel a lot less silly.

Thanks SlipEternal,
That answer is brilliant and I never would of thought of it in 1000 years.
One question though.
$(x-1)(x^4+x^3+x^2+x+1)=x^5-1$

Something in the recesses of my brain tells me I should know this. Could you please refresh my memory on the broader picture of what i am supposed to know here? [(edited) Doesn't matter, I think i got it, I did the second bracket as a sum of a GP. ]

Another unrelated question for SlipEternal.
Anyway. I can't transer the code from Texmaker straight over into here, it is too different. Last time I had to snip the answer that I produced and put it in as an attachment, that wasn't very satisfactory but it was the best I could manage. If I get the one you suggested working will there be less of a problem with this? Do you have any other thoughts on the matter?
Thank you.

5. ## Re: If a then what is b

I need to learn to keep quiet when I'm not 100% certain of something.

6. ## Re: If a then what is b

No I don't think so Romsek.
I think people have to have enough knowledge to know whether to accept an answer or not to accept an answer.
And, even if wrong answers do get accepted occasionally, it is not the end of world.
I am sure at the time you were 100% sure. I came up with the same answer and initially, until I thought more, I was also 100% sure. Don't worry about it and don't start being too hesitant about making posts.
Melody.

7. ## Re: If a then what is b

Exactly, it is the geometric sum:

\begin{align*}(x^n + \cdots + 1)(x-1) & = (x^n + \cdots + 1)x - (x^n + \cdots + 1) \\ & = (x^{n+1} + x^n + \cdots + x) - (x^n + \cdots + x + 1) \\ & = x^{n+1} + (x^n + \cdots + x) - (x^n + \cdots + x) - 1 \\ & = x^{n+1} - 1\end{align*}

As for LaTeX, there is no direct converter that will replace the  from regular LaTeX with  tags. Additionally, many of the formatting features are not available on the forum.

8. ## Re: If a then what is b

Thanks. For what it's worth I believe this sort of thing comes out of the study of rings of polynomials.

9. ## Re: If a then what is b

Originally Posted by SlipEternal
Multiplying out: $x^2+x+2+\dfrac{1}{x} + \dfrac{1}{x^2} = 1$.

Multiply both sides by $x^2$ to get $x^4 + x^3 + x^2 +x + 1 = 0$.

Multiplying both sides by $x-1$ gives $x^5-1 = 0$ (but $x \neq 1$ since we added the root). Hence, $x = -(-1)^{1/5}, x= (-1)^{2/5}, x = -(-1)^{3/5}, x = (-1)^{4/5}$ are the four solutions. In each case, $x^5 = 1$. So:

$\left(x^{20} + \dfrac{1}{x^{20}} + 1\right)\left(x^{20} + \dfrac{1}{x^{20}}\right) = \left((x^5)^4 + \dfrac{1}{(x^5)^4} + 1\right)\left( (x^5)^4 + \dfrac{1}{(x^5)^4}\right) = 3\cdot 2 = 6$
Excellent Analysis...I always believed that in this forum there are excellent mathematicians.. BRAVO..