Exercise 3. Find the integer triplets $(x, y, n)$ such that $n^{2}=17 x^{4}-32 x^{2} y^{2}+41 y^{4}$.
Solution to Exercise 3: We will show by infinite descent that $(0,0,0)$ is the only solution.

Since a square is only 0 or 1 modulo 3 (one way to see this is to consider the cases for the values modulo 3), it is pertinent to attempt a study modulo 3 to see where it leads us. To do this, we note that if $(x, y, n)$ is a solution, then modulo 3, we have

$$
n^{2} \equiv -x^{4} - 2x^{2}y^{2} - y^{4} \equiv -\left(x^{2} + y^{2}\right)^{2} \quad (\bmod 3)
$$

Thus, $n^{2} + (x^{2} + y^{2})^{2}$ is divisible by 3. Since a square is 0 or 1 modulo 3, the only way a sum of squares can be divisible by 3 is if each term is. Therefore, $n$ and $x^{2} + y^{2}$ are divisible by 3. Similarly, it follows that $x$ and $y$ are divisible by 3.

By substituting back into the equation, we deduce that $3^{4} = 81$ divides $n^{2}$ and thus 9 divides $n$, so $\left(\frac{x}{3}, \frac{y}{3}, \frac{n}{9}\right)$ is another solution. If we had a solution with $x, y$, or $n$ non-zero (and by changing them to their opposites, which does not change the values of the squares, which are positive), we could obtain a strictly decreasing sequence of natural numbers, which is absurd (this is the principle of infinite descent).

We conclude that the only potential solution is $(0,0,0)$. Conversely, we note that $(x, y, n) = (0,0,0)$ works (both sides give 0).

The unique solution to the equation is therefore $(0,0,0)$.

Graders' Comments: In general, the solutions that addressed the problem were quite successful. Some attempted approaches such as a study modulo 8 or a rewriting of the equation; here it did not lead to a solution, but these were good ideas that can still be useful on similar problems. We note recurring errors regarding the principle of infinite descent: a number of solutions conclude that there are no solutions by infinite descent. In reality, infinite descent allows us to conclude that there are no non-zero solutions, but it does not exclude the solution $(0,0,0)$ a priori (and indeed it works). On this point, a number of solutions conclude that $(0,0,0)$ is the only possible solution, without verifying it. Even if it seems obvious here, verifying the solution is essential, as infinite descent guarantees that it is the only possible solution, but not that it is indeed a solution.