I know that $a-b+c-1=1$ with $a,b$ and $c$ positive integers. Regrouping the terms we have $a-(b-c+1)=1$. I rewrite that equation as the determinant of a matrix,
$$ e= \begin{pmatrix}{} a & 1 \\ b-c+1 & 1 \end{pmatrix} $$
Now $e$ is in the group $SL(2,\mathbb{Z})$ which is generated by the matrices
$$ S= \begin{pmatrix}{} 0 & -1 \\ 1 & 0 \end{pmatrix} ,\space \space T = \begin{pmatrix}{} 1 & 1 \\ 0 & 1 \end{pmatrix}. $$
These matrices satisfy the relations $S^{4} = (ST)^{6} = 1$, and $S^{2} = (ST)^{3}$
My question is how do I write $e$ as the multiplication of $S$ and $T$? I keep thinking that since $S$ and $T$ are generators of $SL(2,\mathbb{Z})$ then $e$ can be represented (not necessarily uniquely) with the terms $S$ and $T$? Thanks.