I have tried to prove the following theorem.

**Theorem 1.** Let $ q,k \in \mathbb{Z_{\geq 2}}, 1\leq x \leq q-1,$ and $ 0 \leq m \leq k-1.$ Then we have the equality $ $ A(x,q,k,m):=1+qx^{k+m}+\left\lfloor-\frac{qx^{k+m}}{q^k}\right\rfloor+(1-q^k)\left\lfloor \frac{qx^{k+m}}{q^k}+\frac{1}{q^k}\left\lfloor -\frac{qx^{k+m}}{q^k} \right\rfloor \right\rfloor=qx^{k+m} – q^k\left\lfloor\frac{qx^{k+m}}{q^k}\right\rfloor.$ $

Using the inequalities for the floor function I’ve got that $ $ qx^{k+m}-q^k\left\lfloor\frac{qx^{k+m}}{q^k}\right\rfloor-1 \leq A(x,q,k,m) \leq qx^{k+m}-q^k\left\lfloor\frac{qx^{k+m}}{q^k}\right\rfloor. $ $

Is there any way to get rid of that $ -1$ so we would have an equality?