Let $n$ be an integer. Express $A_{n}=a^{n}+a^{-n}$ in terms of $A_{1}=A$ and $n$.

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I. solution. First, express $a$ from the equation

$$
a+a^{-1}=a+\frac{1}{a}=A
$$

which can be rewritten as $a^{2}-A a+1=0$. The product of the two roots is 1, so

$$
a_{1}=\frac{A+\sqrt{A^{2}-4}}{2}, \quad a_{2}=\frac{1}{a_{1}}
$$

and these give only one value for $A_{n}$, since

$$
a_{1}^{n}+\frac{1}{a_{1}^{n}}=\frac{1}{a_{2}^{n}}+a_{2}^{n}
$$

which is

$$
A_{n}=\frac{1}{2^{n}}\left\{\left(A+\sqrt{A^{2}-4}\right)^{n}+\left(A-\sqrt{A^{2}-4}\right)^{n}\right\}
$$

The obtained expression is a polynomial in $A$, because the expansion of the two $n$-th powers only differs in the sign of the odd powers of the square root term, and the sum of these is zero in each pair.

The above statement was intended for positive integer values of $n$. However, since $A_{-n}=A_{n}$ holds for $n$ being a positive integer, and for $n=0$, $A_{0}=2$ for all $a$ (provided $a \neq 0$), and this is also given by (1), (1) is valid for all integer $n$.

For a given $A$, $a$ and $A_{n}$ are real if $|A| \geq 2$. We omit further analysis.

Petz Dénes (Budapest, Veres Pálné Gymn., III. o. t.)

II. solution. Our problem is related to Exercise 1234. There, we derived $A_{n}$ for the values $2 \leq n \leq 7$. Slightly modifying those results, adding the cases for $n=0,1$, and arranging them in ascending powers, we summarize them in the schema of Figure 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_0b0b89a3874697d61324g-1.jpg?height=374&width=657&top_left_y=1320&top_left_x=715)

Figure 1

For these values, we found $A_{n}$ as a polynomial in $A$, with only the exponents congruent to $n$ modulo 2 appearing in each case.

Essentially, we already used the recursion

$$
A_{n+1}=A \cdot A_{n}-A_{n-1}
$$

in the previous derivation, which can be easily verified from the definition of $A_{n}$:

$$
a^{n+1}+\frac{1}{a^{n+1}}=\left(a+\frac{1}{a}\right)\left(a^{n}+\frac{1}{a^{n}}\right)-\left(a^{n-1}+\frac{1}{a^{n-1}}\right)
$$

This proves the validity of the statement about the parity of the exponents for any $n$. Based on this, each coefficient in the polynomial form of $A_{n+1}$ can be derived from the schema in Figure 1 by subtracting the coefficient two rows above from the coefficient directly above it, as indicated by the thick arrows in the schema.

We observe, and it can be easily proven in general, that the sign changes step by step in each column, while it remains constant along the lines sloping to the right (at a $45^{\circ}$ angle). Knowing this, the absolute value of each coefficient is given by the sum of the coefficient directly above it and the coefficient two rows above it.

Papp Zoltán, Göndös Ferenc

Remarks. 1. The method of constructing the absolute values of the coefficients resembles the rule for constructing Pascal's triangle. The placement also matches if we rearrange our schema, writing the coefficients on successive lines along the lines sloping to the right (at a $45^{\circ}$ angle) (Figure 2).

$$
\begin{aligned}
& \begin{array}{l}
2^{2} 3^{1} 5^{2} \\
2{ }^{5} 9^{x^{4}} 5^{5} 1
\end{array} \\
& \text { Figure 2 }
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_0b0b89a3874697d61324g-2.jpg?height=363&width=805&top_left_y=426&top_left_x=649)

This schema, if the internal terms are broken down into sums as in Figure 3, can be considered the sum of two triangular schemas, with the right-hand (smaller digits) schema being Pascal's triangle, and the left-hand (larger digits) schema being twice Pascal's triangle.

The coefficients in the polynomial form of $A_{n}$ can thus be constructed as the sum of twice a binomial coefficient and the next binomial coefficient. For example,

$$
A_{7}=-\left[2 \cdot\binom{3}{2}+\binom{3}{3}\right] A+\left[2 \cdot\binom{4}{1}+\binom{4}{2}\right] A^{3}-\left[2 \cdot\binom{5}{0}+\binom{5}{1}\right] A^{5}+\binom{6}{0} A^{7}
$$

From this, we can read off the general form (it is more convenient to write the terms in reverse order):

$$
\begin{gathered}
A_{n}=\binom{n-1}{0} A^{n}-\left[2 \cdot\binom{n-2}{0}+\binom{n-2}{1}\right] A^{n-2}+\ldots+(-1)^{k}\left[2 \cdot\binom{n-k-1}{k-1}+\right. \\
\left.+\binom{n-k-1}{k}\right] A^{n-2 k}+\ldots
\end{gathered}
$$

This relationship can be proven by the same reasoning, but it is more convenient to prove it directly by induction: based on the above recursion,

$$
\begin{gathered}
A_{n+1}=A A_{n}-A_{n-1}=A \cdot \sum_{k=0}(-1)^{k}\left[2 \cdot\binom{n-k-1}{k-1}+\binom{n-k-1}{k}\right] A^{n-2 k}- \\
-\sum_{j=0}(-1)^{j}\left[2 \cdot\binom{n-1-j-1}{j-1}+\binom{n-1-j-1}{j}\right] A^{n-1-2 j}= \\
=\sum_{k=0}(-1)^{k}\left\{2\left[\binom{n-k-1}{k-1}+\binom{n-k-1}{k-2}\right]+\left[\binom{n-k-1}{k}+\binom{n-k-1}{k-1}\right]\right\} A^{n+1-2 k}= \\
=\sum_{k=0}(-1)^{k}\left[2\binom{n+1-k-1}{k-1}+\binom{n+1-k-1}{k}\right] A^{n+1-2 k} .
\end{gathered}
$$

The binomial coefficients in each coefficient can also be combined:

$$
2\binom{n-k-1}{k-1}+\binom{n-k-1}{k}=\binom{n-k-1}{k-1}\left[2+\frac{n-2 k}{k}\right]=\frac{n}{k}\binom{n-k-1}{k-1}
$$

so

![](https://cdn.mathpix.com/cropped/2024_05_02_0b0b89a3874697d61324g-2.jpg?height=187&width=593&top_left_y=2145&top_left_x=755)

![](https://cdn.mathpix.com/cropped/2024_05_02_0b0b89a3874697d61324g-2.jpg?height=363&width=783&top_left_y=2334&top_left_x=690)

2. Figure 2 can also be interpreted as in Figure 4, with the component written in larger digits appearing by shifting the other one 1 position to the left and 1 position down.

[^0]:    ${ }^{1}$ K. M. L. $39(1969) 59$. o.