6. For a given prime number $p$, determine the number of (all) ordered triples $(a, b, c)$ of numbers $z$ from the set $\left\{1,2,3, \ldots, 2 p^{2}\right\}$, which satisfy the relation

$$
\frac{[a, c]+[b, c]}{a+b}=\frac{p^{2}+1}{p^{2}+2} \cdot c
$$

where $[x, y]$ denotes the least common multiple of numbers $x$ and $y$.
SOLUTION. In the given equation, it will be advantageous to transition from the least common multiples $\mathrm{k}$ to the greatest common divisors, using the known relationship $(x, y) \cdot[x, y]=x \cdot y$ (see problem N3). Let's therefore denote $u=(a, c), v=(b, c)$ and rewrite the left side of the equation as follows:

$$
\frac{[a, c]+[b, c]}{a+b}=\frac{a c / u+b c / v}{a+b}=\left(\frac{a}{u}+\frac{b}{v}\right) \cdot \frac{c}{a+b}
$$

The given equation can therefore (after multiplying by the fraction $(a+b) / c$) be written in the equivalent form

$$
\frac{a}{u}+\frac{b}{v}=\frac{p^{2}+1}{p^{2}+2} \cdot(a+b)
$$

Let's compare the estimates of the expressions in (1). Since $p^{2}>0$, for the fraction on the right side of (1) it is clearly true that

$$
\frac{1}{2}<\frac{p^{2}+1}{p^{2}+2}<1
$$

so as a result of (1), it must be

$$
\frac{a+b}{2}<\frac{a}{u}+\frac{b}{v}<a+b
$$

Due to the left inequality, both natural numbers $u, v$ cannot be greater than 1, because from the inequalities $u \geqq 2$ and $v \geqq 2$ we would get

$$
\frac{a}{u}+\frac{b}{v} \leqq \frac{a+b}{2}
$$

Therefore, at least one of the numbers $u, v$ is equal to one. The right inequality in (2) however excludes the case $u=v=1$. Thus, the number 1 is equal to exactly one of the numbers $u, v$. Considering the symmetry, we will only analyze the case $u=1$ and $v \geqq 2$.

Since we introduced the number $v$ by the relation $v=(b, c)$, the fraction $b / v$ is equal to some natural number $b_{1}$. Let's now substitute the values $u=1$ and $b=b_{1} v$ into (1) and solve the resulting equation with respect to the variable $a$:

$$
\begin{aligned}
a+b_{1} & =\frac{p^{2}+1}{p^{2}+2} \cdot\left(a+b_{1} v\right), \\
\left(p^{2}+2\right)\left(a+b_{1}\right) & =\left(p^{2}+1\right)\left(a+b_{1} v\right), \\
a & =b_{1}\left(\left(p^{2}+1\right) v-p^{2}-2\right) .
\end{aligned}
$$

If it were $v \geqq 3$, we would get from the last equation the estimate

$$
a \geqq\left(p^{2}+1\right) v-p^{2}-2 \geqq 3\left(p^{2}+1\right)-p^{2}-2=2 p^{2}+1
$$

which is in contradiction with the inequality $a \leqq 2 p^{2}$ given by the range in which, according to the problem statement, the values of $a, b, c$ lie. Therefore, the opposite inequality $v<3$ must hold, which together with the assumption $v \geqq 2$ leads to the conclusion that necessarily $v=2$. Equation (3) thus turns into the equation

$$
a=b_{1}\left(2\left(p^{2}+1\right)-p^{2}-2\right)=p^{2} b_{1},
$$

which we can easily solve in the given range of values of $a$, the set $\left\{1,2,3, \ldots, 2 p^{2}\right\}$.

Since $a \leqq 2 p^{2}$, it follows that $b_{1} \leqq 2$. From the conditions $u=(a, c)=1$ and $v=(b, c)=2$ it follows that $c$ is an even number that is coprime with the number $a$. From the equality $a=p^{2} b_{1}$ it follows that $b_{1}=1$ and $p$ is an odd prime. Therefore, $a=p^{2} b_{1}=p^{2}$ and $b=b_{1} v=1 \cdot 2=2$. For the number $c$ this means the following refinement: $c$ is an even number that is not a multiple of the given prime $p$.

Which $c \in\left\{1,2,3, \ldots, 2 p^{2}\right\}$ satisfy this condition and how many of them are there? As we already know, for $p=2$ no such $c$ exists. For odd $p$, from all $p^{2}$ possible even numbers $c=2,4,6, \ldots, 2 p^{2}$ we exclude all multiples of the number $p$, i.e., exactly $p$ numbers $2 p, 4 p, \ldots, 2 p^{2}$; the number of suitable values is therefore precisely $p^{2}-p$. This is the number of all sought triples $(a, b, c)=\left(p^{2}, 2, c\right)$ in the analyzed case, where $u=1$ and $v \geqq 2$. In the second possible case, where on the contrary $v=1$ and $u \geqq 2$, there is, with respect to symmetry, the same number $p^{2}-p$ of suitable triples, which are all of the form $\left(2, p^{2}, c\right)$. (We will include the description of suitable triples in the answer, even though the problem statement does not require it.)

Answer: In the case of $p=2$, no suitable triples exist; in the case of an odd prime $p$, there are exactly $2\left(p^{2}-p\right)$ of them, and all are of the form

$$
(a, b, c)=\left(p^{2}, 2, c\right) \quad \text { or } \quad(a, b, c)=\left(2, p^{2}, c\right)
$$

where $c \in\left\{1,2, \ldots, 2 p^{2}\right\}$ is any even number that is not a multiple of $p$.