Integer solutions to $b^3=(a \bmod b)\cdot(a-a\bmod b)$ Recently, I have found this problem:

Determine all the possible integers solutions $(a,b) \in N_0$ to the following equation with modulus:
  $$b^3=(a \bmod b)\cdot(a-a\bmod b)$$

To solve this, I have thought to use the relation:
$$a \bmod b=a-b\cdot\left \lfloor \frac{a}{b} \right \rfloor$$
Substituting, we have:
$$b^3=\left (a-b\cdot\left \lfloor \frac{a}{b} \right \rfloor\right)\cdot\left(a-\left(a-b\cdot\left \lfloor \frac{a}{b} \right \rfloor\right)\right)$$
Which is the same as:
$$b^3=\left (a-b\cdot\left \lfloor \frac{a}{b} \right \rfloor\right)\cdot b\cdot\left \lfloor \frac{a}{b} \right \rfloor$$
And so:
$$b^3=ab\cdot\left \lfloor \frac{a}{b} \right \rfloor-b^2\cdot\left \lfloor \frac{a}{b} \right \rfloor^2$$
Being $b\neq0$, we can divide and rearrange:
$$b^2=\left \lfloor \frac{a}{b} \right \rfloor\left(\frac{a}{b}-\left \lfloor \frac{a}{b} \right \rfloor\right)$$
Now, we can use the identity:
$$\frac{a}{b}-\left \lfloor \frac{a}{b} \right \rfloor=\left \{  \frac{a}{b}\right \}$$
And so:
$$b=\left \lfloor \frac{a}{b} \right \rfloor\cdot\left \{  \frac{a}{b}\right \}$$
But now, how can we go on? Any idea?

Let $q,r$ be any positive integers such that $b^2=qr$ and $r<b<q$. Then $a=bq+r$ gives a solution: $(a\bmod b)=r$, $a-(a\bmod b)=bq+r-r=bq$, so $(a\bmod b)(a-(a\bmod b))=rbq=b^3$.
For example, let $b=10$, so $b^2=100=20\times5$, so take $q=20$, $r=5$. Then $a=205$, and you can check that $(a\bmod b)(a-(a\bmod b))=(5)(200)=1000=10^3$.
EDIT: Moreover, all solutions are of this form. Given any $a,b$, $b>0$, there exist unique $q,r$ with $a=bq+r$, $0\le r<b$. Then $(a\bmod b)(a-(a\bmod b))=rbq$, so we must have $rbq=b^3$, which is $b^2=qr$. Since $r<b$, we must have $q>b$.
