MHB What Are the Possible Values of the Sum from a 2012-Degree Polynomial Solution?

[magneto1]

Let $x=a$ be a solution of the equation $x^{2012}-7x+6=0$. Find all the possible values for: $1+a+a^2+\cdots+a^{2011}$.

 

[anemone]

Let $x=a$ be a solution of the equation $x^{2012}-7x+6=0$. Find all the possible values for: $1+a+a^2+\cdots+a^{2011}$.

My solution:

Spoiler

We're told $x=a$ is a solution of the equation $x^{2012}-7x+6=0$, therefore we have $a^{2012}-7a+6=0$.

It can be rewritten as

$a^{2012}-1-7a+6+1=0$

$a^{2012}-1-7a+7=0$

$(a^{2012}-1)-7(a-1)=0$

$(a-1)(a^{2011}+a^{2010}+\cdots+a+1)-7(a-1)=0$

$(a-1)(a^{2011}+a^{2010}+\cdots+a+1-7)=0$

So the value of the expression $1+a+a^2+\cdots+a^{2011}$ could be either 2012 or 7.

 

[MarkFL]

My solution:...

Quite clever! (Nod)

 

[magneto1]

My solution:

Spoiler

We're told $x=a$ is a solution of the equation $x^{2012}-7x+6=0$, therefore we have $a^{2012}-7a+6=0$.

It can be rewritten as

$a^{2012}-1-7a+6+1=0$

$a^{2012}-1-7a+7=0$

$(a^{2012}-1)-7(a-1)=0$

$(a-1)(a^{2011}+a^{2010}+\cdots+a+1)-7(a-1)=0$

$(a-1)(a^{2011}+a^{2010}+\cdots+a+1-7)=0$

So the value of the expression $1+a+a^2+\cdots+a^{2011}$ could be either 2012 or 7.

Nicely done that's correct.