Problem 6. For which integers $m$ and $n$ does the equality

$$
(5+3 \sqrt{2})^{m}=(3+5 \sqrt{2})^{n} ?
$$

hold?
Solution. Obviously, the pair $(0 ; 0)$ is a solution to equation (1).

To find other solutions, we use a calculator.

$$
5+3 \sqrt{2} \approx 9.242639 ; 3+5 \sqrt{2}=10.071065
$$

Thus, $3+5 \sqrt{2}>5+3 \sqrt{2}$. Therefore, $n$ and $m$ are either natural numbers or negative. Moreover, if the pair $(n ; m)$ is a solution to equation (1), then the pair $(-n ;-m)$ is also a solution. Therefore, we assume that $n$ and $m$ are natural numbers. Obviously, $n>1.038598 \ldots$.

We continue the search using a calculator: $10: 9>1.038598 \ldots, 11: 10>1.038598 \ldots, 12: 11>$ $>1.038598 \ldots, \ldots, 26: 25>1.038598 \ldots, 27: 26 \approx$ $\approx 1.0384615 \ldots, 28 \cdot 27<1.038598 \ldots$.

Thus, we have a hypothesis that $m=27$ and $n=26$. But how can we justify or refute it?

We need to verify the validity of the equality

$$
\begin{aligned}
&(5+3 \sqrt{2})^{27}=(3+5 \sqrt{2})^{26} \text { or } \\
&\left(\frac{5+3 \sqrt{2}}{3+5 \sqrt{2}}\right)^{26}=\frac{1}{5+3 \sqrt{2}} \text { or } \\
&\left(\frac{(5+3 \sqrt{2})(5 \sqrt{2}-3)}{(5 \sqrt{2}+3)(5 \sqrt{2}-3)}\right)^{26}=\frac{(5-3 \sqrt{2})}{(5+3 \sqrt{2})(5-3 \sqrt{2})} \text { or } \\
&\left(\frac{16 \sqrt{2}+15}{41}\right)^{26}=\frac{5-3 \sqrt{2}}{7}
\end{aligned}
$$

But equality (3) cannot be exact because the coefficient of $\sqrt{2}$ in the left part is positive, while in the right part it is negative.

Now it is clear how to prove that the only integer solution to equation (1) is the pair $(0 ; 0)$. We transform equation (1) as follows:

$$
\begin{aligned}
\left(\frac{5+3 \sqrt{2}}{3+5 \sqrt{2}}\right)^{n} & =\frac{1}{(5+3 \sqrt{2})^{m-n}} \\
\left(\frac{16 \sqrt{2}+15}{41}\right)^{n} & =\left(\frac{5-3 \sqrt{2}}{7}\right)^{m-n}
\end{aligned}
$$

The coefficient of $\sqrt{2}$ in the left part of the last equality is positive, while in the right part it is negative.

## Task 4. FACTORIZATION OF POLYNOMIALS INTO FACTORS