Let the dividend be $a$, the divisor be $b$, the quotient be $q$, and the remainder be $r$, so that
$$a = bq + r,$$
with $0 \leq r < b$.

We are told that $a$ and $b$ are directly proportional to $r$ and $q$, respectively. That is, there exists a constant $k > 0$ such that
$$a = k r, \quad b = k q.$$

Substitute $a$ and $b$ in the division equation:
$$k r = (k q) q + r \implies k r = k q^2 + r.$$
Bring $r$ terms together:
$$k r - r = k q^2 \implies r(k - 1) = k q^2.$$
So
$$r = \frac{k q^2}{k - 1}.$$

But $r$ must be an integer, so $k - 1$ divides $k q^2$. Since $k$ and $k - 1$ are coprime, $k - 1$ must divide $q^2$.

Let $k - 1 = d$, so $k = d + 1$. Then
$$r = \frac{(d + 1) q^2}{d} = q^2 + \frac{q^2}{d}.$$
So $d$ divides $q^2$, i.e., $q^2 = d m$ for some integer $m$.

Thus,
$$r = q^2 + m.$$
But $\frac{q^2}{d} = m$, so $d$ divides $q^2$ and $m$ is an integer.

Recall $b = k q = (d + 1) q$ and $a = k r = (d + 1) r$.

Now, $r$ and $q$ are relatively prime. Since $r = q^2 + m$ and $q^2 = d m$, $r = d m + m = m (d + 1) = m k$.

So $r = m k$, $q^2 = d m$, $k = d + 1$.

But $r$ and $q$ are relatively prime, so $\gcd(m k, q) = 1$.

But $q^2 = d m$, so $q = \sqrt{d m}$. For $q$ to be integer, $d m$ must be a perfect square. Let $d = t^2$, $m = s^2$, so $q = t s$.

Then $r = m k = s^2 (t^2 + 1)$.

Now, $a = k r = (t^2 + 1) \cdot s^2 (t^2 + 1) = s^2 (t^2 + 1)^2$.

Thus, $a$ is a perfect square.

Therefore, the dividend is a perfect square.