3. (USS 3) Find all positive integers $n$ such that
$$
\mathrm{n}=d_{6}^{2}+d_{7}^{2}-1 \text {, }
$$
where $1=d_{1}<d_{2}<\cdots<d_{k}=n$ are all positive divisors of the number n.
3. From the equality $n=d_{6}^{2}+d_{7}^{2}-1$ we see that $d_{6}$ and $d_{7}$ are relatively prime and $d_{7}\left|d_{6}^{2}-1=\left(d_{6}-1\right)\left(d_{6}+1\right), d_{6}\right| d_{7}^{2}-1=\left(d_{7}-1\right)\left(d_{7}+1\right)$. Suppose that $d_{6}=a b, d_{7}=c d$ with $1<a<b, 1<c<d$. Then $n$ has 7 divisors smaller than $d_{7}$, namely $1, a, b, c, d, a b, a c$, which is impossible. Hence, one of the two numbers $d_{6}$ and $d_{7}$ is either a prime $p$ or the square of a prime $p^{2}$, where $p$ is not 2 . Let it be $d_{i}, i \in\{6,7\}$; then $d_{i} \mid\left(d_{j}-1\right)\left(d_{j}+1\right)$ implies that $d_{j} \equiv \pm 1\left(\bmod d_{i}\right)$, and consequently $\left(d_{i}^{2}-1\right) / d_{j} \equiv \pm 1$ as well. But either $d_{j}$ or $\left(d_{i}^{2}-1\right) / d_{j}$ is less than $d_{i}$, and therefore equals $d_{i}-1$. We thus conclude that $d_{7}=d_{6}+1$. Setting $d_{6}=x, d_{7}=x+1$ we obtain that $n=x^{2}+(x+1)^{2}-1=2 x(x+1)$ is even.
(i) Assume that one of $x, x+1$ is a prime $p$. The other one has at most 6 divisors and hence must be of the form $2^{3}, 2^{4}, 2^{5}, 2 q, 2 q^{2}, 4 q$, where $q$ is an odd prime. The numbers $2^{3}$ and $2^{4}$ are easily eliminated, while $2^{5}$ yields the solution $x=31, x+1=32, n=1984$. Also, $2 y$ is eliminated because $n=4 p q$ then has only 4 divisors less than $x ; 2 q^{2}$ is eliminated because $n=4 p q^{2}$ has at least 6 divisors less than $x ; 4 q$ is also eliminated because $n=8 p q$ has 6 divisors less than $x$.
(ii) Assume that one of $x, x+1$ is $p^{2}$. The other one has at most 5 divisors ( $p$ excluded), and hence is of the form $2^{3}, 2^{4}, 2 q$, where $q$ is an odd prime. The number $2^{3}$ yields the solution $x=8, x+1=9, n=144$, while $2^{4}$ is easily eliminated. Also, $2 q$ is eliminated because $n=4 p^{2} q$ has 6 divisors less than $x$.
Thus there are two solutions in total: 144 and 1984 .