# Find the value of the product

#### anemone

##### MHB POTW Director
Staff member
Hey MHB, I am back, fully recover from food poisoning and first off, I want to take this opportunity to wish everyone and their family a very happy and Merry Christmas, much luck, good health and all good things of life. I hope you guys are able to spend it with loved ones!  I want to also present a "Christmas Math Challenge" as a gift to my friends (all of you are my friends, aren't you? ) here today! $a$ satisfies the equation $\left(a+\dfrac{1}{a}+1 \right)\left(a+\dfrac{1}{a} \right)=1$.

What is the value of $\left(a^{20}+\dfrac{1}{a^{20}}+1 \right)\left(a^{20}+\dfrac{1}{a^{20}} \right)$?

#### chisigma

##### Well-known member
Hey MHB, I am back, fully recover from food poisoning and first off, I want to take this opportunity to wish everyone and their family a very happy and Merry Christmas, much luck, good health and all good things of life. I hope you guys are able to spend it with loved ones!  I want to also present a "Christmas Math Challenge" as a gift to my friends (all of you are my friends, aren't you? ) here today! $a$ satisfies the equation $\left(a+\dfrac{1}{a}+1 \right)\left(a+\dfrac{1}{a} \right)=1$.

What is the value of $\left(a^{20}+\dfrac{1}{a^{20}}+1 \right)\left(a^{20}+\dfrac{1}{a^{20}} \right)$?
... with simple steps You find that $\displaystyle a^{4} + a^{3} + a^{2} + a + 1 = 0$, i.e. a must be one of the fifth roots of 1 with the only exclusion of a= 1, so that in any case is $\displaystyle a^{20}=1$... Merry Christmas from Serbia

$\chi$ $\sigma$

#### Klaas van Aarsen

##### MHB Seeker
Staff member On the first day of Christmas
my true love sent to me:
A Partridge in a Pear Tree

On the second day of Christmas
my true love sent to me:
2 Turtle Doves
and a Partridge in a Pear Tree

On the third day of Christmas
my true love sent to me:
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

On the fourth day of Christmas
my true love sent to me:
4 Calling Birds
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

On the fifth day of Christmas
my true love sent to me:
5 Golden Rings
4 Calling Birds
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

On the sixth day of Christmas
my true love sent to me:
6 Geese a Laying
5 Golden Rings
4 Calling Birds
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

Expand to $a^4+a^3+a^2+a+1=0$.
Multiply by $(a-1)$ to give $a^5-1=0$.
So $a^5 = 1$ and therefore $a^{20}=1$ (with $a$ neither rational nor real).
Substitute to find:
$$(1+\frac 1 1 + 1)(1 + \frac 1 1) = 6$$

It's 6 Geese a Laying!

Merry Chrismas!!  #### anemone

##### MHB POTW Director
Staff member
... with simple steps You find that $\displaystyle a^{4} + a^{3} + a^{2} + a + 1 = 0$, i.e. a must be one of the fifth roots of 1 with the only exclusion of a= 1, so that in any case is $\displaystyle a^{20}=1$...

View attachment 1800

Merry Christmas from Serbia

$\chi$ $\sigma$
Thank you chisigma for participating! Your answer is correct!  On the first day of Christmas
my true love sent to me:
A Partridge in a Pear Tree

On the second day of Christmas
my true love sent to me:
2 Turtle Doves
and a Partridge in a Pear Tree

On the third day of Christmas
my true love sent to me:
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

On the fourth day of Christmas
my true love sent to me:
4 Calling Birds
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

On the fifth day of Christmas
my true love sent to me:
5 Golden Rings
4 Calling Birds
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

On the sixth day of Christmas
my true love sent to me:
6 Geese a Laying
5 Golden Rings
4 Calling Birds
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

Expand to $a^4+a^3+a^2+a+1=0$.
Multiply by $(a-1)$ to give $a^5-1=0$.
So $a^5 = 1$ and therefore $a^{20}=1$ (with $a$ neither rational nor real).
Substitute to find:
$$(1+\frac 1 1 + 1)(1 + \frac 1 1) = 6$$

It's 6 Geese a Laying!

Merry Chrismas!!  Hey I like Serena, thanks for the greetings and participating in my challenge problem and the way you solved the problem has opened my eyes to know other method to tackle problem such as this!

My solution:

I noticed

$a^2+\dfrac{1}{a^2} =a^8+\dfrac{1}{a^8}=1-\left( a+\dfrac{1}{a} \right)$,$a^4+\dfrac{1}{a^4}=a^{16}+\dfrac{1}{a^{16}}=\left( a+\dfrac{1}{a} \right)$, $a^{12}+\dfrac{1}{a^{12}}=-1-\left( a+\dfrac{1}{a} \right)$

$\therefore \left( a^4+\dfrac{1}{a^4} \right) \left( a^{16}+\dfrac{1}{a^{16}}\right)=1-\left( a+\dfrac{1}{a} \right)$

and this gives

$a^{20}+\dfrac{1}{a^{20}}+a^{12}+\dfrac{1}{a^{12}}=1-\left( a+\dfrac{1}{a} \right)$

$a^{20}+\dfrac{1}{a^{20}}-1-\left( a+\dfrac{1}{a} \right)=1-\left( a+\dfrac{1}{a} \right)$

$a^{20}+\dfrac{1}{a^{20}}=2$

therefore $\left( a^{20}+\dfrac{1}{a^{20}}+1 \right) \left( a^{20}+\dfrac{1}{a^{20}} \right)=(2+1)(2)=6$