5. Let $T_{k}$ denote the product of the first $k$ odd prime numbers: $T_{1}=3, T_{2}=$ $3 \cdot 5, T_{6}=3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17$ and so on. For each natural number $k$, find the number of natural numbers $n$ such that the number $n^{2}+T_{k} n$ is a perfect square of a natural number. Solve the problem: a) for $k=1$, b) for $k=2$, c) for an arbitrary given natural number $k$.
Answer. $\frac{3^{k}-1}{2}$.

Solution. FIRST METHOD. Let's write the condition of the problem as an equation

$$
n^{2}+T_{k} n=m^{2} \text { for some } m \in \mathbb{N} \text {, }
$$

Let $d=\operatorname{GCD}\left(n, T_{k}\right), \quad \nu=\frac{n}{d}, r=\frac{T_{k}}{d}$. Then the numbers $d, \nu, r$ satisfy the system

$$
\left\{\begin{array}{l}
d^{2} \cdot \nu(\nu+r)=m^{2} \text { for some } m \in \mathbb{N} \\
d r=T_{k}, \\
\text { GCD }(\nu, r)=1 \Leftrightarrow \text { GCD }(\nu, \nu+r)=1
\end{array}\right.
$$

Conversely, each solution $(d, r, \nu)$ of system (2) leads to a solution of equation (1) by the substitution $n=d \nu$. These substitutions are mutually inverse, since in the system (2) the number $d$ turns out to be the greatest common divisor of the numbers $n, T_{k}$. Thus, there is a one-to-one correspondence between the solutions of equation (1) and system (2), i.e., they have the same number of solutions.

From the first and third equalities of system (2), it follows that each of the numbers $\nu, \nu+r$ is a perfect square: $\nu=a^{2}, \nu+r=b^{2}$. Hence it follows

$$
\left\{\begin{array}{l}
d(b-a)(b+a)=T_{k} \\
\operatorname{GCD}(a, b)=1
\end{array}\right.
$$

Conversely, for any $d, a, b$ satisfying (3), the substitution $\nu=a^{2}, r=b^{2}-a^{2}$ leads to a solution of system (2). That is, these systems have the same number of solutions.

Finally, let $b-a=r_{1}, b+a=r_{2}$. Then from system (3) it follows

$$
\left\{\begin{array}{l}
d r_{1} r_{2}=T_{k} \\
r_{1}0 \Rightarrow a>0
$$

Thus, from each $n$ satisfying (5), a pair of natural numbers $a, b$ is obtained, satisfying (6). Let's show that with the inverse substitution, from each such pair $a, b$ a number $n$ is obtained - a solution of equation (5). Let $a, b$ be two numbers satisfying (6). Since $a b=T_{k}^{2} \equiv 1(\bmod 4)$, we have either $a \equiv b \equiv 1(\bmod 4)$ or $a \equiv b \equiv 3(\bmod 4)$. In both cases $a+b \equiv 2 \equiv 2 T_{k}(\bmod 4)$. The inverse substitution is given by the equality

$$
n=\frac{a+b-2 T_{k}}{4}
$$

From the above, it follows that the number $n-$ is an integer. From the inequality $a+b>$ $2 \sqrt{a b}=2 T_{k}$ it follows that $n>0$, i.e., $n \in \mathbb{N}$. By direct substitution, it can be verified that for such $n$ the equality (5) holds if $m=\frac{b-a}{4}$. (From $a \equiv b(\bmod 4)$ it follows that $\frac{b-a}{4}$ is an integer). Finally, for the given product $a b=T_{k}^{2}$ and sum $a+b=4 n+2 T_{k}$, the numbers $a, b$ are uniquely determined, from which it follows that different pairs $(a, b)$ correspond to different $n$.

Thus, there is a one-to-one correspondence between the numbers $n$, satisfying (5), and pairs $(a, b)$, satisfying (6). Let's find the number of such pairs $(a, b)$.

Let $T_{k}^{2}=a b$ be an arbitrary factorization of the number $T_{k}^{2}$. The number of such factorizations is equal to the number of divisors of the number $T_{k}^{2}$, i.e., $3^{k}$. The equality $a=b$ is possible in the unique case $a=b=T_{k}$, in all other $\left(3^{k}-1\right)$ factorizations we have $a \neq b$. The condition $a<b$ is satisfied in half of them, i.e., we have $\frac{3^{k}-1}{2}$ desired pairs $(a, b)$.

Criteria.

(-.) One of the points a), b) is solved (correct answer + full justification).

(-/+) Both points a) and b) are solved, possibly with a minor computational (but not logical) error in point b), not affecting the fundamental correctness of the solution. If one or several cases are missed in the enumeration in point b) - this is a logical error, and the score is no higher than (-.)

(+/2) Both points a), b) are correctly solved and fully justified, and the solution uses an approach (using factorization) that can be generalized for any $k$. An attempt is made to generalize this approach, i.e., the key factorization in general form is written, further progress is not made.

(+/-) Continuation of (+/2): points a) b) are correctly solved, in point c) the problem is clearly reduced to combinatorics, i.e., it is stated: the number of desired numbers $n$ is equal to the number of factorizations of such-and-such a number into such-and-such factors with such-and-such conditions.

(+.) Point c) is solved (the correct answer is presented), but with minor gaps in the justification. For example, the mutual uniqueness of the sets of solutions when reducing the problem to combinatorics is not proven.

(+) Point c) is solved with full justification of all steps, including the one-to-one correspondences between the sets of solutions of the various resulting equations.