For which positive integer $\lambda$ can the equation

$$
x^{2}+y^{2}+z^{2}+v^{2}=\lambda x y z v
$$

be solved in positive integers?
If $\lambda=1$ or $\lambda=4$, then there exists a positive integer solution, for example $x=y=z=v=2$, or $x=y=z=v=1$. We will show that for other $\lambda$ values, there is no positive integer solution.

Suppose for a certain $\lambda$ there is a solution, and consider a solution for which $x+y+z+v$ is minimal. By symmetry, we can assume that $x \leq y \leq z \leq v$.

First, we prove that

$$
v \leq \frac{\lambda x y z}{2} \quad \text{and} \quad v \leq \sqrt{x^{2}+y^{2}+z^{2}}
$$

Consider (1) as a quadratic equation in $v$. According to the relationships between roots and coefficients (Vieta's formulas), the other root is

$$
v_{1}=\lambda x y z - v = \frac{x^{2}+y^{2}+z^{2}}{v}
$$

From the first formula, it is clear that $v_{1}$ is also an integer, and from the second formula, it is clear that $v_{1}$ is positive. This means that (1) also has the solution $(x, y, z, v_{1})$. Since $x+y+z+v$ is minimal, $v \leq v_{1}$, which implies (2) holds by (3).

The following estimate provides an upper bound for $\lambda$:

$$
\lambda x y z = x^{2} + y^{2} + z^{2} + v^{2} \leq x v + y v + z v + v \frac{\lambda x y z}{2} \leq \left(3 + \frac{\lambda}{2}\right) x y z v
$$

From this, $\lambda \leq 3 + \frac{\lambda}{2}$, or $\lambda \leq 6$.

If none of $x, y, z, v$ are even, then the left side is divisible by 4, but the right side $x y z v$ is odd, so $\lambda$ must be divisible by 4; this is only possible if $\lambda=4$.

If one or three of $x, y, z, v$ are even, then the right side is even, but the left side is odd, which is a contradiction.

If two of $x, y, z, v$ are even, then the left side gives a remainder of 2 when divided by 4, but the right side is divisible by 4, which is also a contradiction.

Finally, if all of $x, y, z, v$ are even, then the (4) estimate can be refined slightly:

$$
\lambda x y z v \leq x v + y v + z v + v \frac{\lambda x y z}{2} \leq \left(\frac{3}{4} + \frac{\lambda}{2}\right) x y z v; \quad \text{from this, } \lambda \leq \frac{3}{4} + \frac{\lambda}{2}, \text{ or } \lambda=1
$$

Therefore, a positive integer solution is only possible for $\lambda=1$ and $\lambda=4$.

Remarks. 1. According to (3), for both $\lambda$ values, infinitely many solutions can be constructed, since one of $x, y, z, v$ - for example, the smallest one - can be replaced by a larger integer.

2. A very similar equation $\left(x^{2} + y^{2} + z^{2} = k x y z\right)$ is examined in the book "Selected Problems and Theorems in Elementary Mathematics" by Jaglom-Skljarszkij-Csencov (Volume 1, Problem 118, Tankönyvkiadó, 1967), and the solutions of the two problems are essentially the same. Many have rewritten the proof described there for equation (1). Unfortunately, the solution in the book contains several inaccuracies, and some contestants did not correct them. Many, following the book, defined the number $v_{1}$ by the formula $v_{1} = \lambda x y z - v$ and verified by substitution that the quadruple $(x, y, z, v_{1})$ is also a solution to (1). However - like in the book - they forgot to justify why $v_{1}$ is positive. (These contestants received 4 points.) Two even got entangled in the same copying mistake, as they copied the most critical sentences word for word from the book.