Transform the following sum into a product:

$$
x^{3n+2} + x^{3k+1} + 1
$$

where $n$ and $k$ are natural numbers.
I. Solution: Let us denote our expression by $P(x)$, then (if $n \geq k$)

\[
\begin{aligned}
& (x-1) P(x)=\left(x^{3 n+2}+x^{3 k+1}+1\right)(x-1)= \\
= & \left(x^{3 n+3}-1\right)-\left(x^{3 n+2}-x^{3 k+2}\right)-\left(x^{3 k+1}-x\right)= \\
= & {\left[\left(x^{3}\right)^{n+1}-1\right]-x^{3 k+2}\left[\left(x^{3}\right)^{n-k}-1\right]-x\left[\left(x^{3}\right)^{k}-1\right]=} \\
= & \left(x^{3}-1\right)\left[\left(x^{3}\right)^{n}+\left(x^{3}\right)^{n-1}+\ldots+\left(x^{3}\right)+1\right]- \\
- & x^{3 k+2}\left[\left(x^{3}-1\right)^{n-k-1}+\left(x^{3}\right)^{n-k-2}+\ldots+\left(x^{3}\right)+1\right]-x\left(x^{3}-1\right)\left[\left(x^{3}\right)^{k-1}+\ldots+1\right]
\end{aligned}
\]

From each of the three terms, $x^{3}-1$ can be factored out, so

\[
(x-1) P(x)=\left(x^{3}-1\right) P_{1}(x)
\]

where $P_{1}(x)$ is a rational integer function of $x$. From (1),

\[
P(x)=\frac{x^{3}-1}{x-1} P_{1}(x)=\left(x^{2}+x+1\right) P_{1}(x)
\]

If $n<k$, then from the second term of the second row, we do not factor out the $x^{3 k+2}$ term, but the $-x^{3 n+2}$ term, and thus the second term of the third row becomes

\[
-x^{3 n+2}\left[\left(x^{3}\right)^{k-n}-1\right]
\]

The further transformation remains the same, only $n$ and $k$ are swapped in the second term of $P_{1}(x)$.

Gergely Ervin (Bp. IV., Konyves Kalman g. IV. o. t.)

II. Solution: Our expression can also be written as

\[
\begin{aligned}
& x^{3 n+2}-x^{2}+x^{3 k+1}-x+x^{2}+x+1=x^{2}\left[\left(x^{3}\right)^{n}-1\right]+x\left[\left(x^{3}\right)^{k}-1\right]+\left(x^{2}+x+1\right)= \\
& =x^{2}\left(x^{3}-1\right)\left(x^{3(n-1)}+x^{3(n-2)}+\ldots+1\right)+x\left(x^{3}-1\right)\left(x^{3(k-1)}+x^{3(k-2)}+\ldots+1\right)+\left(x^{2}+x+1\right)= \\
& =\left(x^{2}+x+1\right)\left[x^{2}(x-1)\left(x^{3(n-1)}+x^{3(n-2)}+\ldots+1\right)+x(x-1)\left(x^{3(k-1)}+x^{3(k-2)}+\ldots 1\right)+1\right]
\end{aligned}
\]

Stahl Janos (Bp. VI., Kolcsey g. IV. o. t.)

Note: Of course, a factorization was not acceptable in which not both factors were rational expressions in $x$.