# Factorize

1. Sep 2, 2010

### The legend

1. The problem statement, all variables and given/known data

Factor the equation (without complex numbers)

$$a^{10} + a^{5} + 1$$

3. The attempt at a solution

I substituted $$a^{5} = x$$ getting a quadratic eqation. But when I factored the quadratic equation I get complex roots and this is against the question.

2. Sep 2, 2010

### iamalexalright

x^2 + x + 1

I don't know if this is the final answer they are looking for but you could add a 'clever' version of 0 to simplify it a bit

3. Sep 2, 2010

### Dickfore

The solutions of $x^{2} + x + 1 = 0$ are complex numbers:

$$x_{1/2} = \frac{-1 \pm i \sqrt{3}}{2}$$

The modulus of these numbers is:

$$|x_{1/2}| = 1$$

and their arguments (limited to the interval $[0, 2\pi)$) are:

$$\arg{(x_{1})} = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$

$$\arg{(x_{2})} = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$$

Take the fifth root of each of them and combine the complex conjugate roots in five pairs.

4. Sep 3, 2010

### The legend

Sorry, but i dont know this part of mathematics. Can you please suggest some place(website etc) where i can learn it by myself?

5. Sep 3, 2010

### Dickfore

How do you expect to go to a Mathematics Olympiad and don't know about complex numbers?

6. Sep 4, 2010

### The legend

complex numbers are not in the syllabus of the olympiad. Thats why i mentioned it in the question.

7. Sep 4, 2010

### The legend

I tried out a different approach(without complex nos) and got it!

Take $$a^{10} + a^{5} + 1 = y$$ and $$a^5 = x$$

then

$$y = x^{2} + x + 1$$
$$(x - 1) y = (x - 1)(x^{2} + x + 1)$$
$$(x - 1) y = x^{3} - 1$$
$$y = \frac{a^(5)3 - 1^{5}}{a^{5} - 1}$$

Now simplifying using (x^5 - y^5) and dividing i get the answer to be

$$y = (a^{2} + a + 1)(a^{8} - a^{7} + a^{5} -a^{4} + a^{3} - a + 1)$$

Right , isnt it?

anyway..Dickfore, can you please suggest a good book or website to self learn complex numbers?