Find the number of real solutions to
\[(x^{2006} + 1)(x^{2004} + x^{2002} + x^{2000} + \dots + x^2 + 1) = 2006x^{2005}.\]
Note that $x = 0$ is not a solution.  Also, if $x < 0,$ then the left-hand side is positive and the right-hand side is negative, so $x$ cannot be a solution.  Thus, any real roots must be positive.  Assume $x > 0.$

Dividing both sides by $x^{2005},$ we get
\[\frac{(x^{2006} + 1)(x^{2004} + x^{2002} + x^{2000} + \dots + x^2 + 1)}{x^{2005}} = 2006.\]Then
\[\frac{x^{2006} + 1}{x^{1003}} \cdot \frac{x^{2004} + x^{2002} + x^{2000} + \dots + x^2 + 1}{x^{1002}} = 2006,\]or
\[\left( x^{1003} + \frac{1}{x^{1003}} \right) \left( x^{1002} + x^{1000} + x^{998} + \dots + \frac{1}{x^{998}} + \frac{1}{x^{1000}} + \frac{1}{x^{1002}} \right) = 2006.\]By AM-GM,
\begin{align*}
 x^{1003} + \frac{1}{x^{1003}} &\ge 2, \\
x^{1002} + x^{1000} + x^{998} + \dots + \frac{1}{x^{998}} + \frac{1}{x^{1000}} + \frac{1}{x^{1002}} &\ge \sqrt[1003]{x^{1002} \cdot x^{1000} \cdot x^{998} \dotsm \frac{1}{x^{998}} \cdot \frac{1}{x^{1000}} \cdot \frac{1}{x^{1002}}} = 1003,
\end{align*}so
\[\left( x^{1003} + \frac{1}{x^{1003}} \right) \left( x^{1002} + x^{1000} + x^{998} + \dots + \frac{1}{x^{998}} + \frac{1}{x^{1000}} + \frac{1}{x^{1002}} \right) \ge 2006.\]Since we have the equality case, the only possible value of $x$ is 1, so there is $\boxed{1}$ real root.
1
Level 4
Intermediate Algebra
Problem: Find the number of real solutions to
\[(x^{2006} + 1)(x^{2004} + x^{2002} + x^{2000} + \dots + x^2 + 1) = 2006x^{2005}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the number of real solutions, we first analyze the equation given. The left-hand side consists of two factors: \(x^{2006} + 1\) and \(x^{2004} + x^{2002} + x^{2000} + \dots + x^2 + 1\). The term \(x^{2006} + 1\) is always positive for all real \(x\) since \(x^{2006}\) is non-negative and adding 1 ensures positivity.

Next, we simplify the second factor. The expression \(x^{2004} + x^{2002} + x^{2000} + \dots + x^2 + 1\) can be recognized as a geometric series with \(n = 1003\) terms, where the first term is \(1\) and the common ratio is \(x^2\). The sum of this series can be expressed as:

\[
\frac{x^{2006} - 1}{x^2 - 1} \quad \text{for } x^2 \neq 1.
\]

This series is also positive for all real \(x\) except at \(x = 1\) and \(x = -1\), where it equals \(1003\).

Now, we equate the left-hand side to the right-hand side, \(2006x^{2005}\). The right-hand side is a polynomial of degree 2005, which can have at most 2005 real roots. 

Next, we analyze the behavior of the left-hand side. As \(x\) approaches positive or negative infinity, \(x^{2006}\) dominates, and thus the left-hand side approaches infinity. At \(x = 0\), the left-hand side equals \(1\), while the right-hand side equals \(0\). 

We check specific values:
- At \(x = 1\), the left-hand side equals \(2007\) and the right-hand side equals \(2006\).
- At \(x = -1\), the left-hand side equals \(2007\) and the right-hand side equals \(-2006\).

The left-hand side is continuous and differentiable, and it transitions from \(1\) at \(x = 0\) to larger values as \(x\) increases. The right-hand side, being a polynomial, is also continuous. 

By the Intermediate Value Theorem, there must be at least one intersection point between the two sides in the interval \((0, 1)\) since the left-hand side starts below the right-hand side at \(x = 0\) and exceeds it at \(x = 1\). 

To find the total number of solutions, we consider the derivatives. The left-hand side is a product of two positive functions, and its derivative will be positive for \(x > 0\), indicating that it is strictly increasing. The right-hand side, being a polynomial of degree 2005, can have multiple turning points but is also increasing for large \(x\).

Given the analysis, we conclude that there is exactly one point where the two sides intersect, leading to the final answer of \(1\) real solution.<|END_OF_LATENT|> The final answer is $1$.