Exercise 1. Determine all pairs of integers $(x, y)$ such that $x^{2}+73=y^{2}$.
Solution to Exercise 1: As is often the case with a Diophantine equation, we seek to rearrange the equation so as to have products of factors on both sides of the equality. When we are dealing with perfect squares, we can use the remarkable identity $y^{2}-x^{2}=(y-x)(y+x)$. This allows us to rewrite the equation as:

$$
73=(y-x)(y+x)
$$

Thus, the number $y-x$ is a divisor of 73. The divisors of 73 are $73, 1, -1, -73$. We then distinguish four cases:

Case $\mathrm{n}^{\circ} 1: y-x=73$. Then $x+y=1$. This leads to $\mathrm{y}=73+x=73+(1-\mathrm{y})$, so $\mathrm{y}=37$ and $x=-36$. Conversely, we indeed have $(-36)^{2}+73=37^{2}$, so the pair $(-36,37)$ is indeed a solution.

Case $\mathrm{n}^{\circ} 2: y-x=1$. Then $x+y=73$. This leads to $y=1+x=1+(73-y)$, so $y=37$ and $x=36$. Conversely, we indeed have $36^{2}+73=37^{2}$, so the pair $(36,37)$ is indeed a solution.

$\underline{\text { Case } \mathrm{n}^{\circ} 3: y}-\mathrm{x}=-1$. Then $x+y=-73$. This leads to $y=-1+x=-1+(-73-y)$, so $y=-37$ and $x=-36$. Conversely, we indeed have $(-36)^{2}+73=(-37)^{2}$, so the pair $(-36,-37)$ is indeed a solution.

Case $\mathrm{n}^{\circ} 4: y-x=-73$. Then $x+y=-1$. This leads to $y=-73+x=-73+(-1-y)$, so $y=-37$ and $x=36$. Conversely, we indeed have $36^{2}+73=(-37)^{2}$, so the pair $(36,-37)$ is indeed a solution.

The solution pairs are therefore $(-36,-37),(-36,37),(36,-37)$ and $(36,37)$.

## Comments from the graders:

Exercise well done overall, with some recurring errors:

- Assuming the integers are positive.
- Forgetting cases in the factorization (forgetting to take the opposites / to swap the factors)