IMO 1996 SL N4

Origin: BUL | Category: Number Theory

Problem

Find all positive integers $a$ and $b$ for which

$$\left[\frac{a^2}{b}\right] + \left[\frac{b^2}{a}\right] = \left[\frac{a^2 + b^2}{ab}\right] + ab$$

where as usual, $[t]$ refers to greatest integer that is less than or equal to $t$.

Solution

Let $a, b \in \mathbb{N}$ satisfy the given equation. It is not possible that $a = b$ (since it leads to $a^2 + 2 = 2a$), so we assume w.l.o.g. that $a > b$. Next, for $a > b = 1$ the equation becomes $a^2 = 2a$, and one obtains a solution $(a, b) = (2, 1)$.

Let $b > 1$. If

$$\left[\frac{a^2}{b}\right] = \alpha \quad \text{and} \quad \left[\frac{b^2}{a}\right] = \beta,$$

then we trivially have $ab \geq \alpha\beta$. Since also

$$\left[\frac{a^2+b^2}{ab}\right] \geq 2,$$

we obtain $\alpha + \beta \geq \alpha\beta + 2$, or equivalently

$$(\alpha - 1)(\beta - 1) \leq -1.$$

But $\alpha \geq 1$, and therefore $\beta = 0$. It follows that $a > b^2$, i.e., $a = b^2 + c$ for some $c > 0$. Now the given equation becomes

$$b^3 + 2bc + \left[\frac{c^2}{b}\right] = \left[\frac{b^4+2b^2c+b^2+c^2}{b^3+bc}\right] + b^3 + bc,$$

which reduces to

$$(c - 1)b + \left[\frac{c^2}{b}\right] = \left[\frac{b^2(c + 1) + c^2}{b^3 + bc}\right]. \tag{1}$$

If $c = 1$, then $(1)$ always holds, since both sides are $0$. We obtain a family of solutions $(a, b) = (n, n^2 + 1)$ or $(a, b) = (n^2 + 1, n)$. Note that the solution $(1, 2)$ found earlier is obtained for $n = 1$.

If $c > 1$, then $(1)$ implies that

$$\left[\frac{b^2(c+1)+c^2}{b^3+bc}\right] \geq (c - 1)b.$$

This simplifies to

$$c^2(b^2 - 1) + b^2(c(b^2 - 2) -(b^2 + 1)) \leq 0. \tag{2}$$

Since $c \geq 2$ and $b^2 - 2 \geq 0$, the only possibility is $b = 2$. But then $(2)$ becomes

$$3c^2 + 8c - 20 \leq 0,$$

which does not hold for $c \geq 2$.

Hence the only solutions are $(n, n^2 + 1)$ and $(n^2 + 1, n)$, $n \in \mathbb{N}$.