I found that there is not an odd number that is providing this.

Because if $k$ is odd $a$ and $b$ have to be square numbers. Let $a=x^2$ and $b=y^2$. And $2a+3b$ have to be a square number. Let $2a+3b=z^2$. Then $2x^2+3y^2=z^2$ $3y^2\equiv0(mod 3)$. $2x^2\equiv{z^2}(mod 3)$ $x\equiv0(mod 3)$ and $z\equiv0(mod 3)$ Then let $z=3m$ and $x=3k$ $18k^2+3y^2=9m^2$ $6k^2+y^2=3m^2$ $y\equiv0(mod 3)$ Let $y=3n$ $6k^2+9n^2=3m^2$ $2k^2+3n^2=m^2$ That is Fermat method. There is not any solution if $k$ is odd.

If $k$ is even, $(a,b)=(2p^{k/2-1},3p^{k/2-1})$ satisfies the condition for each prime $p>13$.

kaede wrote: If $k$ is even, $(a,b)=(2p^{k/2-1},3p^{k/2-1})$ satisfies the condition for each prime $p>13$. But how? pls explain

MATHS_wizard33 wrote: But how? pls explain Use this \begin{align*} n&=p_1^{\alpha _1}p_2^{\alpha _2}\cdots p_m^{\alpha _m}\\ s(n)&=\prod_{i=1}^{m}(\alpha _i+1) \end{align*}