x + 1/x = 3, y la escalera hasta x⁵ + 1/x⁵

Give me one fact, that $x + \tfrac{1}{x} = 3$, and I can hand you $x^5 + \tfrac{1}{x^5}$ without ever finding $x$. That still feels a bit like cheating the first time you see it. The two values of $x$ that fit the condition are ugly, both stuffed with $\sqrt{5}$, yet the thing we actually want comes out a clean whole number: $123$. No decimals, no radicals left over. This is the engine behind a whole genre of competition problems, and it runs on a two-line rule you can do on your fingers.

Turn the condition into two roots

Multiply $x + \tfrac{1}{x} = 3$ through by $x$ and you get a plain quadratic:

Here is the move that makes everything else easy. If $x$ solves this, so does $\tfrac{1}{x}$, because the product of the two roots is $1$ (the constant term over the leading one). So $x$ and $\tfrac{1}{x}$ are the two roots of the same quadratic, and Vieta’s formulas tell us all we will ever need about them:

Sum $3$, product $1$. We are never going to solve for $x$. We are going to build everything out of those two numbers.

Name what you want, then find its rule

Give the quantity a name. Let

so the target is $p_5$. Two floors are free: $p_0 = x^0 + x^0 = 2$ and $p_1 = 3$, which is the condition itself. Now watch what happens when you multiply the given sum by one of these:

The cross terms collapse: $x^{n-1}\cdot \tfrac{1}{x} = x^{n-2}$ and $\tfrac{1}{x^{n-1}}\cdot x = \tfrac{1}{x^{n-2}}$. The left side is just $3\,p_{n-1}$, so rearranging gives the whole trick in one line:

Read equation (2) slowly, because it is the reason the radicals vanish. Every term is three times the last one minus the one before, and that is nothing but whole-number arithmetic. Start from two integers, $2$ and $3$, and you can never leave the integers. The $\sqrt{5}$ baked into $x$ quietly cancels with the $\sqrt{5}$ baked into $\tfrac{1}{x}$ at every single step.

Climb to the fifth floor

With the rule in hand there is no cleverness left, only bookkeeping. Each line uses the two above it:

The same ladder answers the question for any exponent. Want $x^{10} + \tfrac{1}{x^{10}}$? Keep climbing: $p_6 = 322$, and so on. One rule, as many floors as you like.

This is Newton’s identities in miniature

Equation (2) is not a coincidence of this particular puzzle. The numbers $p_n$ are the power sums of the two roots $x, \tfrac{1}{x}$, and their elementary symmetric functions are exactly the sum and product from Vieta: $e_1 = x + \tfrac{1}{x} = 3$ and $e_2 = x\cdot\tfrac{1}{x} = 1$. Newton’s identities say that for two variables,

which for $e_1 = 3,\ e_2 = 1$ is our equation (2) on the nose. Change the condition to $x + \tfrac{1}{x} = s$ and the recurrence just swaps the $3$ for $s$: $p_n = s\,p_{n-1} - p_{n-2}$. That single formula quietly solves every “find $x^k + 1/x^k$” problem you will ever be handed.

The number hiding under 123

We refused to solve for $x$, but let us peek now. The quadratic (1) gives

The larger root is $\tfrac{3+\sqrt5}{2} \approx 2.618$, and that is a familiar face in disguise. Writing the golden ratio as $\varphi = \tfrac{1+\sqrt5}{2}$, a one-line check using $\varphi^2 = \varphi + 1$ gives $\varphi^2 + \varphi^{-2} = 3$. So our $x$ is simply $\varphi^2$, and

where $L_m$ are the Lucas numbers $2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, \dots$ Take every other one and you get our climb exactly:

That sequence $2, 3, 7, 18, 47, 123, 322, \dots$ is catalogued as OEIS A005248, the even-indexed Lucas numbers, and it lists the very recurrence we derived, $a_n = 3a_{n-1} - a_{n-2}$. So the answer to the puzzle is $L_{10} = 123$. The ladder that dodged every square root was counting powers of the golden ratio the whole way up. I find that a little startling: a throwaway algebra drill and one of the most over-discussed constants in mathematics turn out to be the same object, seen from two angles.