6. Find all natural numbers $n$ such that the number $n^{2}+77 n$ is a perfect square of a natural number.
Answer: 4, 99, 175, 1444.

Solution. FIRST METHOD. Let $n(n+77)=m^{2}$, where $m, n$ are natural numbers. Consider 4 cases:

- $\gcd(n, 77)=1$. In this case, the numbers $n$ and $n+77$ are coprime, and thus each of them is a perfect square: $n=p^{2}, n+77=q^{2}$. Hence, $q^{2}-p^{2}=77$, i.e., either $q-p=1, q+p=77$, or $q-p=7, q+p=11$. In the first case, we get $q=39, p=38, n=1444$, in the second case $q=9, p=2, n=4$.
- $\gcd(n, 77)=7$, i.e., $n=7s$, where $s$ does not divide 11. Then $49s(s+11)=m^{2}$. Since the numbers $s$ and $s+11$ are coprime, each of them is a perfect square: $s=p^{2}, s+11=q^{2}$. Hence, $q^{2}-p^{2}=11$, i.e., $q-p=1, q+p=11, q=6, p=5, s=25, n=175$.
- $\gcd(n, 77)=11$, i.e., $n=11s$, where $s$ does not divide 7. Then $121s(s+7)=m^{2}$. Since the numbers $s$ and $s+7$ are coprime, each of them is a perfect square: $s=p^{2}, s+7=q^{2}$. Hence, $q^{2}-p^{2}=7$, i.e., $q-p=1, q+p=7, q=4, p=3, s=9, n=99$.
- $\gcd(n, 77)=77$, i.e., $n=77s$. Then $77^{2}s(s+1)=m^{2}$. Since the numbers $s$ and $s+1$ are coprime, each of them is a perfect square: $s=p^{2}, s+1=q^{2}$. Hence, $q^{2}-p^{2}=(q-p)(q+p)=1$, which means $p=0$, i.e., $s=n=0$. Contradiction, since by condition $n$ is a natural number.

SECOND METHOD. Let $n(n+77)=m^{2}$, where $m, n$ are natural numbers. Multiplying both sides by 4 and completing the left side to a perfect square, we get

$$
\begin{gathered}
4 n^{2}+4 \cdot n \cdot 77+77^{2}=4 m^{2}+77^{2} \\
(2 n+77)^{2}=4 m^{2}+77^{2} \\
(2 n+77-m)(2 n+77+m)=77^{2}
\end{gathered}
$$

Let $2 n+77-m=a, 2 n+77+m=b$. Then

$$
\begin{gathered}
a b=77^{2}, a<b, \\
2 n+77=\frac{a+b}{2}, n=\frac{a+b-2 \cdot 77}{4}
\end{gathered}
$$

From condition (1) for the pair $(a, b)$, we have 4 cases: $\left(1,77^{2}\right),\left(7,7 \cdot 11^{2}\right),\left(11,11 \cdot 7^{2}\right),(49,121)$. For each of them, we compute $n$ according to condition (2), obtaining the same answers as above.

Criteria.

$(-)$ Answer without any attempt at proof.

(-.) Only the case where both numbers $n$ and $n+77$ are perfect squares is considered.
$(-/+)$ One value of $n$ is missing.

(+/2) A solution similar to the first author's solution, but it is not proven that $77 / \gcd(n,77)$ can be represented as the difference of perfect squares.

(+) All four values of $n$ are correctly found, and it is proven that there are no other solutions.