Prove that for infinitely many pairs $(a,b)$ of integers the equation \[x^{2012} = ax + b\] has among its solutions two distinct real numbers whose product is 1.
Problem
Source: 2012 Baltic Way, Problem 4
Tags: quadratics, algebra, quadratic formula, algebra unsolved
nolispe
22.11.2012 21:45
nsato wrote: Prove that for infinitely many pairs $(a,b)$ of integers the equation \[x^{2012} = ax + b\] has among its solutions two distinct real numbers whose product is 1. I'm not sure if this solution would get full credit because I didn't back-substitute my answer at the end and prove that it satisfies the equation. But all my steps are reversible, so I believe it should work.
Let $P(x) = x^{2012} - ax - b$, and suppose $(a, b)$ is a pair of integers such that $P(k) = P(1/k) = 0$.
$k^{2012} = ka + b$
$\frac{1}{k^{2012}} = \frac{a}{k} + b$
Multiplying these two equations together gives;
$1 = (ka+b)\left(\frac{a}{k} + b\right)$
$k = (ka+b)(a + bk)$
$k = ka^2 + k^2ab + ab + kb^2$
Writing the right hand side as a quadratic in terms of $k$ gives:
$0 = abk^2 + (a^2 + b^2 - 1)k + ab$
And using the quadratic formula we find that:
$k = \frac{-(a^2+b^2 - 1) \pm \sqrt{(a^2+b^2 - 1)^2 - (2ab)^2}}{2ab}$
$= \frac{-(a^2+b^2 - 1) \pm \sqrt{((a+b)^2-1)((a-b)^2-1)}}{2ab}$
Thus, any pair of integers $(a, b)$ yielding a positive discriminant will result in a real value for $k$. There are infinite number of such solutions, but since we need distinct values of $k$, we should have $D \neq 0$ as well, so $a+b \neq \pm 1$ and $a-b \neq \pm 1$. For example, the solution set $\{(a, 2-a) | a\geq 3\}$ will work. Then $D = 3(4a^2-8a+3)$, and the condition that $a\geq 3$ guarantees that the discriminant is positive.
mahanmath
30.11.2012 20:15
If you come closer you'll see the connection of these type of equation with recursive sequence.
Start with a trinomial equation $P(x) \in \mathbb Z$ with $P(0)=1$ and two real roots, namely $r_1, r_2$.
Then clearly $r_1 . r_2 =1$
For example start with $P(x)=x^2 - 3x +1$ .Pick up one of roots,say $r$, according to this relation we can compute $r^{2012}$ in this way :
$r^{2012} = (r^2)^{1006} = (3r-1)^{1006} = \dots$
Whenever you have $r^n$ for $n \geq 2$ you can break it to $(r^2) (r^{n-2})$ and replace $r^2$ with $3r-1$ .
So finally you reach a integer coefficients linear value for $r^{2012}$ in terms of $r$ like $ar+b$.
Uagu
22.08.2013 12:03
nolispe wrote: nsato wrote: Prove that for infinitely many pairs $(a,b)$ of integers the equation \[x^{2012} = ax + b\] has among its solutions two distinct real numbers whose product is 1. But all my steps are reversible, so I believe it should work.
Let $P(x) = x^{2012} - ax - b$, and suppose $(a, b)$ is a pair of integers such that $P(k) = P(1/k) = 0$.
$k^{2012} = ka + b$
$\frac{1}{k^{2012}} = \frac{a}{k} + b$
Multiplying these two equations together gives;
$1 = (ka+b)\left(\frac{a}{k} + b\right)$
$k = (ka+b)(a + bk)$
$k = ka^2 + k^2ab + ab + kb^2$
Writing the right hand side as a quadratic in terms of $k$ gives:
$0 = abk^2 + (a^2 + b^2 - 1)k + ab$
And using the quadratic formula we find that:
$k = \frac{-(a^2+b^2 - 1) \pm \sqrt{(a^2+b^2 - 1)^2 - (2ab)^2}}{2ab}$
$= \frac{-(a^2+b^2 - 1) \pm \sqrt{((a+b)^2-1)((a-b)^2-1)}}{2ab}$
Thus, any pair of integers $(a, b)$ yielding a positive discriminant will result in a real value for $k$. There are infinite number of such solutions, but since we need distinct values of $k$, we should have $D \neq 0$ as well, so $a+b \neq \pm 1$ and $a-b \neq \pm 1$. For example, the solution set $\{(a, 2-a) | a\geq 3\}$ will work. Then $D = 3(4a^2-8a+3)$, and the condition that $a\geq 3$ guarantees that the discriminant is positive.
It seems that you have got something wrong, nolispe. A positive discriminant is only a necessary condition for the equation set. However, $1 = (ka+b)\left(\frac{a}{k} + b\right)$ doesn't mean that both $k^{2012} = ka + b$ and $\frac{1}{k^{2012}} = \frac{a}{k} + b$ are true, because we don't know whether the value of $ka+b$ equals $k^{2012}$ or not. So actually not all your steps are reversible.
xxp2000
24.08.2013 03:46
$(n\pm\sqrt{n^2-1})^{2012}=\alpha_n\pm\beta_n\sqrt{n^2-1}$ So $x^{2012}=\beta_nx+(\alpha_n-n\beta_n)$ has two roots $n\pm\sqrt{n^2-1}$.