8.3 The number $\sqrt{1+2019^{2}+\frac{2019^{2}}{2020^{2}}}+\frac{2019}{2020}$ is an integer. Find it.
Solution. Let $a=2020$. Then the desired number is $\sqrt{1+(a-1)^{2}+\frac{(a-1)^{2}}{a^{2}}}+$ $\frac{a-1}{a}$. We will perform equivalent transformations:

$$
\begin{gathered}
\sqrt{1+(a-1)^{2}+\frac{(a-1)^{2}}{a^{2}}}+\frac{a-1}{a}=\frac{\sqrt{a^{2}(a-1)^{2}+(a-1)^{2}+a^{2}}}{a}+\frac{a-1}{a}= \\
=\frac{\sqrt{a^{4}-2 a^{3}+3 a^{2}-2 a+1}}{a}+\frac{a-1}{a}=\frac{\sqrt{\left(a^{2}-a+1\right)^{2}}}{a}+\frac{a-1}{a}=\frac{a^{2}-a+1+a-1}{a}=a
\end{gathered}
$$

Thus, the given number is 2020.

Answer: 2020.

| THERE IS IN THE SOLUTION | SCORE |
| :--- | :---: |
| Correct and justified answer | 7 points |
| Calculations are done "head-on", with arithmetic errors | deduct 1 point for each error |
| In the correct solution, formulas not included in the school curriculum are used but not proven (e.g., $\left.\sqrt{a^{2}+(a+1)^{2}+a^{2} \cdot(a+1)^{2}}=a^{2}+a+1\right)$ | 5 points |
| The problem is reduced to an algebraic one by a successful substitution, but there are errors in further transformations | 3 points |
| Correct answer without justification or with incorrect justification | 0 points |