Problem 1445

We use the well known fact that given a triple $(a, b, c)$ of integers satisfying $a^2 + b^2 = c^2$, then one of $a, b$ is divisible by $4$; one of $a, b$ is divisible by $3$; and one of $a, b, c$ is divisible by $5$. Thus $4, 3$ and $5$ are divisors of $n$. Let us write
$$n = 2^a 3^b 5^c t$$
where $\alpha \ge 2$ and $t$ is not divisible by $2, 3, 5$. Note that $\beta \ge 1, \gamma \ge 1$.
Hence
$$d_1 = 1,\ d_2 = 2,\ d_3 = 3,\ d_4 = 4,\ d_5 = 5,\ d_6 = 6.$$
Moreover $10 = 2 \cdot 5$ is also a divisor of $n$. Hence $d_7 \le 10$. Thus $d_7 \in \{7, 8, 9, 10\}$. We explore each of them.

Case 1: Suppose $d_7 = 7$. In this case
$$(d_{16} - d_{15}) (d_{16} + d_{15}) = d_{16}^2 - d_{15}^2 = d_7^2 = 49.$$
It follows that $d_{16} - d_{15} = 1$ and $d_{16} + d_{15} = 49$. We thus get $d_{16} = 25$, $d_{15} = 24$. Since $d_{15}|n$, we see that $8$ divides $n$. Similarly $25 = 5^2$ also divides $n$. It follows that $\alpha \ge 3, \gamma \ge 2$. Since $d_7 = 7$ is also a divisor of $n$, we may now write
$$n = 2^a 3^b 5^c 7^d t_1$$
where $t_i$ is not divisible by $2, 3, 5, 7$; $\alpha \ge 3, \beta \ge 1, \gamma \ge 2, \delta \ge 1$.

If $\beta \ge 2$, then $9$ divides $n$ and hence $d_1 = 1, d_2 = 2, d_3 = 3, d_4 = 4, d_5 = 5, d_6 = 6, d_7 = 7, d_8 = 8, d_9 = 9, d_{10} = 10, d_{11} \le 12, d_{12} \le 14, d_{13} \le 15, d_{14} \le 18, d_{15} \le 20$. This contradicts $d_{15} = 24$. We conclude that $\beta = 1$ and hence $n = 2^a \cdot 3 \cdot 5^\gamma \cdot 7^\delta t_1$.

If $\alpha \ge 4$, then we have
$d_1 = 1, d_2 = 2, d_3 = 3, d_4 = 4, d_5 = 5, d_6 = 6, d_7 = 7, d_8 = 8, d_9 = 10, d_{10} \le 12, d_{11} \le 14, d_{12} \le 15, d_{13} \le 16, d_{14} \le 20, d_{15} \le 21$, which again contradicts $d_{15} = 24$. This shows that $\alpha = 3$ and hence $n = 2^3 3 \cdot 5^7 7^\delta t_1$ where $\text{gcd}(t_1, 210) = 1$.

Thus we get
$d_1 = 1, d_2 = 2, d_3 = 3, d_4 = 4, d_5 = 5, d_6 = 6, d_7 = 7, d_8 = 8, d_9 = 10, d_{10} \le 12, d_{11} \le 14, d_{12} \le 15, d_{13} \le 20, d_{14} \le 21, d_{15} \le 24$.
Since $d_{15} = 24$ and $d_{16} = 25$, $t_1$ is not divisible by $11, 13, 17, 19$ or $23$. Otherwise $d_{15} < 24$. This shows that $\gamma \ge 2$ and $d_{17}$ cannot be equal to $26 = 2 \cdot 13$ nor be equal to $27 = 3^3$. Since $4$ and $7$ are divisors of $n$, it follows that $28 = 4 \cdot 7$ also divides $n$. Hence $d_{17} = 28$.

Case 2: Suppose $d_7 = 8$. Then
$$(d_{16} - d_{15}) (d_{16} + d_{15}) = 64 = 2 \times 32 = 4 \times 16$$
so that $d_{16} = 17, d_{15} = 15$ or $d_{16} = 10, d_{15} = 6$. Since $d_6 = 6$, we can immediately rule out $d_{15} = 6$. If $d_7 = 8$ and $d_{15} = 15$, then
$$8 = d_7 < d_8 < d_9 < d_{10} < d_{11} < d_{12} < d_{13} < d_{14} < d_{15} = 15,$$
which is impossible.

Case 3: If $d_7 = 9$, then
$$(d_{16} - d_{15}) (d_{16} + d_{15}) = 81 = 1 \times 81 = 3 \times 27.$$
We get $d_{16} = 41, d_{15} = 40$ or $d_{16} = 15, d_{15} = 12$.
Since
$$9 = d_7 < d_8 < \dots < d_{15},$$
we may rule out $d_{15} = 12$. If $d_{15} = 40$, then $40$ divides $n$ and hence $8$ also divides $n$. But $d_6 = 6$ and $d_7 = 9$ and $8$ cannot be a divisor of $n$.

Case 4: Suppose $d_7 = 10$. In this case
$$(d_{16} - d_{15}) (d_{16} + d_{15}) = 100 = 2 \times 50$$
and hence $d_{16} = 26, d_{15} = 24$. This shows $8|n$. But then $d_6 = 6, d_{10} = 10$ is impossible. We conclude that the only possible value is $d_{17} = 28$.