It is given that $ a^2 + a + 1 = 0 $, so what will be the value of $a^{2023} + a^{2021} + 1$?

Question

Given that

$$ a^2 + a + 1 = 0, $$

find the value of

$$ a^{2023} + a^{2021} + 1. $$

I come up with a solution $a^3 = 1$ , but if I put $a = 1$ then the answer changes.

Answer 1

First we can rewrite the equation as

$a^2+1=-a$

Then in the expression, $a^{2023} +a^{2021}+1=0$, we can rewrite that as $a^{2021}(a^{2}+1)+1 \implies a^{2021}(-a)+1$ [from the first equation]

Now the above expression simplifies to $-a^{2022}+1$.

Now the roots of the equation $a^2+a+1=0$ are $w$ and $w^2$.

Thus lets assume $a=w$ and we know that $w^{3n}=1$ where n is any positive integer.

Now we notice 2022 is divisible by 3, so $a^{2022}=w^{2022}=1$.

Thus our answer will be $-1+1=0$

Answer 2

In this answer, I want to talking about the point that confuses you .

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$a^2+a+1=0$ implies that, $a^3=1$ .

Because, $a^2+a+1=0$ simply implies $a≠1$ . Therefore, multiplying both side of the equation by $(a-1)$, you get $a^3=1$. But, recall that we already stated $a≠1$ .

To summarize :

The statement $a^2+a+1=0$ is equivalent to :

$$a^3=1 \thinspace \thinspace \thinspace \color{red}{\text {and}} \thinspace \thinspace \thinspace a≠1\thinspace .$$

But,

The statement $a^2+a+1=0$ is not equivalent to :

$$a^3=1 \thinspace .$$

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The reverse statement is simply incorrect : $a^3=1$ doesn't imply $a=1$ . Because, by the Fundamental Theorem of Algebra $a^3=1$ has $3$ complex roots . Indeed,

$$ \begin{align}&a^3=1\\ \iff &a^3-1=0\\ \iff &(a-1)(a^2+a+1)=0\\ \iff &a=1\thinspace \thinspace \thinspace \color{red}{\text{or}}\thinspace \thinspace \thinspace \underbrace {a^2+a+1=0}_{\Delta_a<0, \thinspace a\thinspace \not\in\thinspace \mathbb R}\end{align} $$

Therefore, you correctly showed that $a^3=1$. But, this doesn't imply $a=1$ . Thus,

$$\begin{align}a^3=1\implies a^{2022}&=\left(a^3\right)^{674}=1\end{align}$$

This leads to , $$a^{2021}(a^2+a+1)=0\thinspace .$$

Answer 3

Let's begin by writing: $$a^{2023}+a^{2021}+1=a^{2021}(a^2+1)+1$$ From the first equation we have $a^2+a+1=0 \Rightarrow a^2+1=-a$ so \begin{align} a^{2023}+a^{2021}+1&=a^{2021}(a^2+1)+1 \\ &=a^{2021}(-a)+1 \\ &=-a^{2022}+1 \end{align} From the first equation we have again \begin{align} a^2+a+1&=0 \\ (a-1)(a^2+a+1)&=0 \\ a^3-1&=0 \\ a^3 &= 1 \end{align} Then note that $a^{2022}=(a^3)^{667}$ so we can have \begin{align} a^{2023}+a^{2021}+1&=-a^{2022}+1 \\ &=-(a^3)^{667}+1 \\ &=-(1^{667})+1 \\ &=-1+1=0 \end{align} Hence $a^{2023}+a^{2021}+1=0$

Answer 4

Note that $2023 \equiv 1\pmod 3$, so we can write $2023 = 3j+1$ for some $j \in \mathbb{N}$; similarly $2021 \equiv 2\pmod 3$, so we can write $2023 = 3k+2$ for some $k \in \mathbb{N}$. Then, using the fact you determined that $a^3=1$,

$$\begin{align} a^{2023} + a^{2021} + 1 &= a^{3j+1} + a^{3k+2} + 1 \\[2mm] &= \left( a^3 \right)^j a^1 + \left( a^3 \right)^k a^2 + 1 \\[2mm] &= a + a^2 + 1 \\[2mm] &= 0 \end{align}$$