245. One angle of the triangle is twice as large as another. Find the sides of the triangle, knowing that they are expressed as integers.
245. $A=2B$. Let $BC=a, AC=b, AB=c$. Obviously $a>b$.

Draw the bisector $AD$ of angle $A$, then $\angle CDA = \angle DAB + \angle DBA = 2B; \angle CDA = \angle CAB$. By the property of the angle bisector in a triangle, as well as considering the similarity of triangles, we find $c=\frac{a^2-b^2}{b}$ (1). From (1) and the condition $ca-b$, it is satisfied since $\frac{a^2-b^2}{b}=\frac{a+b}{b}(a-b)>a-b$.

Thus, $1<\frac{a}{b}<2$ (2).

Equation (1) can be solved in integers as follows: $\frac{a}{b}=\frac{b}{a}+\frac{c}{a}$, let $\frac{a}{b}=\frac{m}{n}$, then $\frac{m}{n}=\frac{n}{m}+\frac{c}{a}, \frac{c}{a}=\frac{m^2-n^2}{mn}$; thus, $\frac{b}{a}=\frac{n}{m}=\frac{n^2}{mn}$, $\frac{c}{a}=\frac{m^2-n^2}{mn}$; therefore, the sides of the triangle are determined in integers by the formulas:

$$
a=mn, b=n^2, c=m^2-n^2
$$

To satisfy condition (2), it is necessary that $m$ and $n$ satisfy the condition $n<m<2n$.

Note. The equation $a^2=b(b+c)$ can be obtained by a trigonometric method, namely: $\frac{a}{b}=\frac{\sin 2B}{\sin B}$, hence $\cos B=\frac{a}{2b}$; but by the cosine theorem $\cos B=\frac{a^2+c^2-b^2}{2ac}$, therefore, $\frac{a}{2b}=\frac{a^2+c^2-b^2}{2ac}$, from which $a^2=b(b+c)$. From the equality $\cos B=\frac{a}{2b}$ we find $a<2b$.