Exercise 6. Find all pairs $(x, y)$ of strictly positive integers such that $x y \mid x^{2}+2 y-1$.
Solution to Exercise 6: On the one hand, we can write $x | xy | x^2 + 2y - 1$ so $x | 2y - 1$. Therefore, there exists $n$ such that $2y - 1 = nx$. Necessarily, $n$ and $x$ are odd.

On the other hand, the divisibility relation in the statement gives us an inequality: we know that $xy > 0$, so

$$
xy \leqslant x^2 + 2y - 1
$$

By multiplying this relation by 2 to simplify the calculations and combining the two, we get

$$
x(nx + 1) \leqslant 2x^2 + 2(nx + 1) - 2
$$

Rearranging the terms,

$$
(n-2)x^2 \leqslant (2n-1)x
$$

Since $x > 0$,

$$
x \leqslant \frac{2n-1}{n-2} = 2 + \frac{3}{n-2}
$$

Remembering that $n$ and $x$ are necessarily odd, we distinguish several cases.

Case $n^{\circ} 1: n \geqslant 7$. Then

$$
x \leqslant 2 + \frac{3}{n-2} < 3
$$

So $x = 1$. Conversely, all pairs $(1, y)$ are solutions.

Case $n^{\circ} 2: n = 5$. Then

$$
x \leqslant 2 + \frac{3}{n-2} = 3
$$

So $x = 1$ or $x = 3$. All pairs with $x = 1$ are solutions. If $x = 3$, we have $2y - 1 = nx = 5 \cdot 3 = 15$ so $y = 8$. We verify that $(3, 8)$ is a solution: $24 \mid 24$.

Case $n^{\circ} 3: n = 3$. Then

$$
x \leqslant 2 + \frac{3}{n-2} = 5
$$

So $x = 1$ or $x = 3$. All pairs with $x = 1$ are solutions. If $x = 3$, we have $2y - 1 = nx = 3 \cdot 3 = 9$ so $y = 5$. We verify that $(3, 5)$ is not a solution: $15 \nmid 24$. If $x = 5$ we have $2y - 1 = nx = 5 \cdot 3 = 15$ so $y = 8$. We verify that $(5, 8)$ is a solution: $40 \mid 40$.

Case $n^{\circ} 4: n = 1$. Then $x = 2y - 1$. Then

$$
xy = 2y^2 - y \mid 4y^2 - 2y = (2y - 1)^2 + 2y - 1 = x^2 + 2y + 1
$$

So all pairs of the form $(2y - 1, y)$ are solutions.

## Comments from the graders:

The exercise is generally well done by the students who tackled it. Despite the somewhat tedious disjunction of cases, there are very few omissions. Some still have difficulty with the concept of divisibility, particularly how to interpret an expression of the form $a \mid bc$.