11.4. How many Pythagorean triangles exist with one of the legs equal to 2013? (A Pythagorean triangle is a right triangle with integer sides. Equal triangles are counted as one.).
Answer: 13. Solution. From the Pythagorean theorem, we obtain the equation in integers $2013^{2}+x^{2}=y^{2} \Leftrightarrow(y-x)(y+x)=2013^{2}=3^{2} \cdot 11^{2} \cdot 61^{2}$. This equation is equivalent to the system $\left\{\begin{array}{l}y-x=d_{1} \\ y+x=d_{2}\end{array}\right.$, where $d_{1}, d_{2}=\frac{2013^{2}}{d_{1}}-$ are natural divisors of the number 2013 2. The solution of the system $x=\frac{d_{2}-d_{1}}{2}, y=\frac{d_{2}+d_{1}}{2}$ will be integer, since $d_{1}$ and $d_{2}-$ are odd numbers. For the number $x$ to be natural, the divisor $d_{2}$ must be greater than $d_{1}$, i.e. $d_{2}>2013, d_{1}<2013$ (we discard the "central" divisor $d=2013$, and the remaining divisors are divided into pairs $\left(d_{1}, d_{2}\right)$, where $d_{1} \cdot d_{2}=2013$, and from each pair we choose $\left.d_{2}>d_{1}\right)$.