Transform the following expression into a product:

$$
x^{7}+x^{6} y+x^{5} y^{2}+x^{4} y^{3}+x^{3} y^{4}+x^{2} y^{5}+x y^{6}+y^{7}
$$
I. Solution: Divide the 8 terms into 4 consecutive term pairs. Each such pair can be written as a product, one of whose factors is $x+y$, because the second term of each such pair is derived from the first by replacing a single $x$ factor with $y$, while the other factors are common. Thus, $x+y$ is a factor of the entire expression:

$$
K=x^{6}(x+y)+x^{4} y^{2}(x+y)+x^{2} y^{4}(x+y)+y^{6}(x+y)=(x+y)\left(x^{6}+x^{4} y^{2}+x^{2} y^{4}+y^{6}\right)
$$

Here, the second polynomial can similarly be transformed, using $x^{2}$ and $y^{2}$ instead of $x$ and $y$, and

$$
K=(x+y)\left(x^{2}+y^{2}\right)\left(x^{4}+y^{4}\right)
$$

This completes the solution of the problem.

Rády Erika (Szeged, Rózsa F. g. II. o. t.)

II. Solution: Recognizing that $K$ has the same structure as the longer polynomial in the factorization of the difference of two equal powers, we can write $K=\left(x^{8}-y^{8}\right) /(x-y)$, of course assuming that $x \neq y$. Now, considering the numerator as the difference of two squares, and then applying this idea to the first factor, and finally using the division possibility,

$$
\begin{gathered}
K=\frac{\left(x^{4}-y^{4}\right)\left(x^{4}+y^{4}\right)}{x-y}=\frac{\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right)\left(x^{4}+y^{4}\right)}{x-y}= \\
=(x+y)\left(x^{2}+y^{2}\right)\left(x^{4}+y^{4}\right)
\end{gathered}
$$

Gyaraki Károly (Tata, Eötvös J. g. I. o. t.)

Remarks. 1. The property mentioned in the first observation of the I. solution holds for any two consecutive terms of $K$, so $K$ can also be considered as the sum of a geometric series with a common ratio of $y / x$. From this, using the geometric series sum formula, we can also arrive at the starting point of the II. solution, provided that $x \neq 0$ and $y / x \neq 1$, i.e., $y \neq x$.

Horváth Tibor (Szombathely, Nagy Lajos g. II. o. t.)

2. The basis for the agreement mentioned in the 1st remark is that the geometric series sum formula is merely another form of the factorization of the difference of two equal powers.

3. We can also arrive at the above factorization (considering the polynomial as a product of linear factors) using Bézout's theorem (see high school IV. o. textbook).

Gáti Pál (Pécs, Nagy Lajos g. II. o. t.)

4. By breaking down the pleasant symmetry of form (1), the factor $x^{4}+y^{4}$ can be further factored:

$$
x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=\left(x^{2}+y^{2}-\sqrt{2} x y\right)\left(x^{2}+y^{2}+\sqrt{2} x y\right)
$$

However, the two resulting factors, as well as the factor $x^{2}+y^{2}$, cannot be factored further in real numbers, because if we consider $x$ as the unknown and $y$ as a constant, and each factor as the left side of a reduced quadratic equation, the discriminant of these equations is negative: $-2 y^{2}$, or $-2 y^{2}$, or $-4 y^{2}$, meaning that the equations $x^{2}-\sqrt{2} y x+y^{2}=0$, $x^{2}+\sqrt{2} y x+y^{2}=0$, $x^{2}+y^{2}=0$ have no real roots.

Fritz József (Mosonmagyaróvár, Kossuth L. g. II. o. t.)

5. The pairing method in the I. solution could be continued until the last factor also became a binomial, because $K$ has 8 terms, which is $2^{3}$. It can be shown by complete induction that with 7 replaced by $2^{k}-1$,

$$
\begin{gathered}
x^{2^{k}-1}+x^{2^{k}-2} y+x^{2^{k}-3} y^{2}+\ldots+y^{2^{k}-1}=(x+y)\left(x^{2}+y^{2}\right)\left(x^{2^{2}}+\right. \\
\left.+y^{2^{2}}\right) \ldots\left(x^{2^{k-1}}+y^{2^{k-1}}\right) \quad 1
\end{gathered}
$$

Kéry Gerzson (Sopron, Széchenyi I. g. I. o. t.)[^0]


[^0]:    ${ }^{1}$ With $y=1$, this is the 1st problem of the 1941 Eötvös competition.