23rd Junior Turkish Mathematical Olympiad

Answer: All positive even integers. For $k=2m$ where $m\in {\mathbb{Z}}^{+}$, $a=2\cdot 5^{m-1}$ and $b=3\cdot 5^{m-1}$ satisfy the required equalities.

Now suppose that for some odd positive integer $k$ there exist such $a$ and $b$. Since $\sigma (n)$ is odd if and only if $n$ is a perfect square, there exist positive integers $x,y$ and $z$ such that $a=x^2$, $b=y^2$ and $2a+3b=2x^2+3y^2=z^2$. Let $\gcd (x,y,z)=t$. Then $x=t{x}_1$, $y=t{y}_1$ and $z=t{z}_1$ where $\gcd (x_1,y_1,z_1)=1$ and hence $2x_1^2+3y_1^2=z_1^2$. Considering this equation in (mod 8) gives us that each of $x_1,y_1$ and $z_1$ is even which is a contradiction. Therefore, for any odd positive integer $k$ there exist no such $a$ and $b$.