Let $\omega$ be a complex number such that $\omega^7 = 1$ and $\omega \neq 1$. Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Then $\alpha$ and $\beta$ are roots of the quadratic
\[x^2 + px + q = 0\]
for some integers $p$ and $q$. Find the ordered pair $(p,q)$.
can 't read it, don't use codes, they don't work here
the actual answer is here (it says i cant post urls... so just add "h t t p s" to the beginning of the phrase below):
://math.stackexchange.com/questions/1253515/quadratic-using-the-roots-of-unity-where-omega7-1-omega-neq-1
Well, let's think about this. We know that $\omega^7 = 1$, which means that $\omega^7 - 1 = 0$. Now, we can factor this as a difference of cubes:
\[(\omega - 1)(\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1) = 0.\]
Since we're given that $\omega \neq 1$, we know that the second factor must equal $0$.
Now, let's substitute $\alpha$ and $\beta$ into this equation, and see what happens:
\[(\alpha - 1)(\alpha^6 + \alpha^5 + \alpha^4 + \alpha^3 + \alpha^2 + \alpha + 1) = 0,\]
and
\[(\beta - 1)(\beta^6 + \beta^5 + \beta^4 + \beta^3 + \beta^2 + \beta + 1) = 0.\]
See what I did there? I just replaced $\omega$ with $\alpha$ and $\beta$! So now we just need to simplify these expressions.
Since $\alpha$ equals $\omega + \omega^2 + \omega^4$, we can factor out $\omega$:
\begin{align*}
\alpha &= \omega(1 + \omega + \omega^3) \\
&= \omega(1 + \omega + \omega^2 + \omega^3 + \omega^4 - \omega^2) \\
&= \omega(1 + \omega + \omega^2)(1 + \omega^3 - \omega^2).
\end{align*}
Similarly, we can factor $\beta$:
\begin{align*}
\beta &= \omega^3(1 + \omega^2 + \omega^4) \\
&= \omega^3(1 + \omega^2 + \omega^3 + \omega^4 + \omega^5 - \omega^5) \\
&= \omega^3(1 + \omega + \omega^2)(1 + \omega^4 - \omega^5).
\end{align*}
Now, let's plug these back into our equations:
\begin{align*}
(\alpha - 1)(\alpha^6 + \alpha^5 + \alpha^4 + \alpha^3 + \alpha^2 + \alpha + 1) &= 0 \\
\left(\omega(1 + \omega + \omega^2)(1 + \omega^3 - \omega^2) - 1\right)\left(\left(\omega(1 + \omega + \omega^2)(1 + \omega^3 - \omega^2)\right)^6 + \left(\omega(1 + \omega + \omega^2)(1 + \omega^3 - \omega^2)\right)^5 + \cdots + \left(\omega(1 + \omega + \omega^2)(1 + \omega^3 - \omega^2)\right) + 1\right) &= 0.
\end{align*}
Now, let's do the same thing for $\beta$:
\begin{align*}
(\beta - 1)(\beta^6 + \beta^5 + \beta^4 + \beta^3 + \beta^2 + \beta + 1) &= 0 \\
\left(\omega^3(1 + \omega + \omega^2)(1 + \omega^4 - \omega^5) - 1\right)\left(\left(\omega^3(1 + \omega + \omega^2)(1 + \omega^4 - \omega^5)\right)^6 + \left(\omega^3(1 + \omega + \omega^2)(1 + \omega^4 - \omega^5)\right)^5 + \cdots + \left(\omega^3(1 + \omega + \omega^2)(1 + \omega^4 - \omega^5)\right) + 1\right) &= 0.
\end{align*}
So, we can see that both $\alpha$ and $\beta$ are roots of quadratic equations of the form $x^2 + px + q = 0$. Now, let's focus on the coefficient of the $x$ term. In both equations, the $x$ term is obtained by multiplying two terms together -- namely, $\alpha$ and $\beta$. So, let's look at the product of $\alpha$ and $\beta$:
\begin{align*}
\alpha \cdot \beta &= \omega(1 + \omega + \omega^2)(1 + \omega^3 - \omega^2) \cdot \omega^3(1 + \omega + \omega^2)(1 + \omega^4 - \omega^5) \\
&= \omega^4(1 + \omega + \omega^2)^2(1 + \omega^3 - \omega^2)(1 + \omega^4 - \omega^5).
\end{align*}
Now, let's simplify this a bit. Notice that $1 + \omega + \omega^2$ is actually equal to $1 - \omega^3$. Why? Well, you can actually check this by multiplying everything out. Similarly, $1 + \omega^3 - \omega^2$ can be written as $1 - \omega^4$. So, plugging this into our equation, we get:
\[\alpha \cdot \beta = \omega^4(1 - \omega^3)^3(1 - \omega^4)(1 + \omega^4 - \omega^5).\]
Now, let's simplify this equation some more. Notice that $1 - \omega^3$ is just the negative of $\omega^3 - 1$. Similarly, $1 - \omega^4$ is just the negative of $\omega^4 - 1$. So, we can write our equation as:
\begin{align*}
\alpha \cdot \beta &= \omega^4(-1)^3(\omega^3 - 1)^3(-1)(\omega^5 - \omega^4 + 1) \\
&= -\omega\left(\omega^3 - 1\right)^3\left(\omega^5 - \omega^4 + 1\right) \\
&= -\omega^9 + 3\omega^6 - 3\omega^3 + 1.
\end{align*}
Finally, let's compare our equation to the standard quadratic form $x^2 + px + q = 0$. The coefficient of the $x$ term is $p$, and the constant term is $q$. So, in our case, we can see that $p = 3$ and $q = 1$. Therefore, the ordered pair $(p, q)$ is $(3, 1)$.
To find the values of $\alpha$ and $\beta$, we can substitute the given expressions into the quadratic equation $x^2 + px + q = 0$ and see what values of $p$ and $q$ make it true.
First, let's calculate the values of $\alpha$ and $\beta$ by plugging in the expressions for $\omega$:
$\alpha = \omega + \omega^2 + \omega^4$
Using the fact that $\omega^7 = 1$, we can simplify this expression:
$\alpha = \omega + \omega^2 + \omega^4 = \omega + \omega^2 + \omega^4 + \omega^7 = \omega + \omega^2 + \omega^4 + \omega = \omega + \omega^2 + \omega^4 + \omega^3$
Now, notice that $\omega + \omega^2 + \omega^3 + \omega^4 = \omega(1 + \omega + \omega^2 + \omega^3) = \omega(1 + \omega(1 + \omega + \omega^2)) = \omega(1 + \omega(1 + \omega(1 + \omega)))$
Since $\omega^7 = 1$, we can substitute $\omega^7$ with 1 to get:
$\alpha = \omega(1 + \omega(1 + \omega(1 + \omega))) = \omega(1 + \omega(1 + \omega(1 + \omega^7))) = \omega(1 + \omega(1 + \omega(1 + 1))) = \omega(1 + \omega(1 + \omega(2))) = \omega(1 + \omega(1 + 2\omega))$
Next, let's simplify the expression for $\beta$:
$\beta = \omega^3 + \omega^5 + \omega^6$
Using the fact that $\omega^7 = 1$, we can also simplify this expression:
$\beta = \omega^3 + \omega^5 + \omega^6 = \omega^3 + \omega^5 + \omega^6 + \omega^7 = \omega^3 + \omega^5 + \omega^6 + \omega = \omega^3 + \omega^5 + \omega^6 + \omega^2$
Similar to what we did for $\alpha$, notice that $\omega^3 + \omega^4 + \omega^5 + \omega^6 = \omega^3(1 + \omega + \omega^2 + \omega^3) = \omega^3(1 + \omega(1 + \omega + \omega^2)) = \omega^3(1 + \omega(1 + \omega(1 + \omega)))$
Substituting $\omega^7$ with 1, we get:
$\beta = \omega^3(1 + \omega(1 + \omega(1 + \omega))) = \omega^3(1 + \omega(1 + \omega(1 + \omega^7))) = \omega^3(1 + \omega(1 + \omega(1 + 1))) = \omega^3(1 + \omega(1 + \omega(2))) = \omega^3(1 + \omega(1 + 2\omega))$
Now, we can rewrite the equation $x^2 + px + q = 0$ by substituting $\alpha$ and $\beta$:
$(x - \alpha)(x - \beta) = 0$
Expanding this expression, we get:
$x^2 - (\alpha + \beta)x + \alpha\beta = 0$
Comparing this with the original equation, we can see that $p = -(\alpha + \beta)$ and $q = \alpha\beta$.
Therefore, the ordered pair $(p,q)$ is $(-(\alpha + \beta), \alpha\beta)$.
This problem is discussed here:
http://math.stackexchange.com/questions/2144834/let-alpha-omega-omega2-omega4-and-beta-omega3-omega5
google is your friend. Try using it.