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How many monic quadratic equations are there, which remain unchanged when both of their roots are squared, and what are those equations? (You must consider the set of all complex numbers, not just real numbers..... :P)

Note by Abhimanyu Swami
4 years, 2 months ago

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Not entirely sure, but here is my work so far:

$$(x - a)(x - b) = (x - a^2)(x - b^2)$$. Expanding, we get $$x^2 - (a + b)x + ab = x^2 - (a^2 + b^2) x + a^2b^2$$, so $$a + b = a^2 + b^2$$ and $$ab = a^2b^2$$. We know trivially the case where $$a$$ or $$b$$ are equal to $$0$$ or $$1$$, so henceforth assume $$a, b \neq 0$$. Now we can divide the second equation to get $$ab = 1$$.

$a + b = a^2 + b^2$ $ab = 1$

Substituting, this is $$a + \frac{1}{a} = a^2 + \frac{1}{a^2} \rightarrow a^4 - a^3 - a - 1 = 0$$. We can factor by grouping: $$a^3(a - 1) - 1(a - 1) \rightarrow (a^3 - 1)(a - 1) \rightarrow (a - 1)^2(a^2 + a + 1) = 0$$. As found before, $$a = 1$$ and now, $$a = \frac{-1 \pm i\sqrt{3}}{2}$$. Since the system of equations are symmetric with respect to $$a$$ and $$b$$, we know that $$b$$ is just the other value (i.e. if $$a$$ takes the "plus," then $$b$$ takes the minus, and vice versa). So, in summary:

$$(a, b): (0, 0) (0, 1), (1, 1), (\frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2})$$, which yields the quadratics $$x^2, x^2 - x, x^2 - 2x - 1, x^2 + x + 1$$.

Alternatively, we have two cases:

One, each of the roots when squared yields the same number. i.e., the roots $$a, b$$ satisfy $$a^2 = a$$ and $$b^2 = b$$. This, of course, occurs at $$0$$ and $$1$$.

Two, one of the roots squared yields the other root, and the other root squared equals the first root, i.e. $$a^2 = b$$ and $$b^2 = a$$. Thus, $$a^4 - a = 0 \rightarrow a(a - 1)(a^2 + a + 1)$$ and we come to the same conclusion as before.

This is under the assumption that the polynomials are monic. Otherwise, just multiply each polynomial by a constant C to get the whole family of quadratics.

- 4 years, 2 months ago

Wow................both of your solutions are elegant and creative. I couldn't think of that. Is there any way to solve problem using Vieta's formula?

- 4 years, 2 months ago

Well, my first solution was essentially using vieta's formulas.

- 4 years, 2 months ago

Your second solution is rather creative. I like it.

- 4 years, 2 months ago

Thanks. Another cool thing is that they are the two complex roots of unity for $$x^3 = 1$$. So essentially what we could have solved instead was the following:

$$a \equiv 2b \pmod k$$

$$b \equiv 2a \pmod k$$

For $$a, b \neq k$$. And apparently, there is only a solution for $$k = 3$$.

For clarity, let $$\mathbb{U}_n^q$$ be the $$qth$$ root of unity for $$x^n = 1$$, otherwise stated as $$\cos \frac{q \pi}{n} + i \sin \frac{q \pi}{n}$$. Then, when we square $$\mathbb{U}_n^q$$ we get $$\mathbb{U}_n^{2q \pmod 7}$$. Going back to the question at hand, if the two roots of the quadratic are $$\mathbb{U}_n^a$$ and $$\mathbb{U}_n^b$$, then we require that $$a \equiv 2b \pmod n$$ and $$b \equiv 2a \pmod n$$.

- 4 years, 2 months ago

Technically, and infinite amount. For example, all those in the form $$z x^2$$. You should specify that the quadratic must be monic.

- 4 years, 2 months ago

Well, I meant monic quadratic. Sorry I didn't specify it. I've edited my post.

- 4 years, 2 months ago