Peter wrote: Let $ n$ be a positive integer with $ k\ge22$ divisors $ 1 = d_{1} < d_{2} < \cdots < d_{k} = n$, all different. Determine all $ n$ such that \[ {d_{7}}^{2} + {d_{10}}^{2} = \left( \frac {n}{d_{22}}\right)^{2}. \] Obviosly $ d_{10}<\frac{n}{d_{22}}=d_{k+1-22}<d_{10}\sqrt 2.$ Because for all solutions $ x^2+y^2=z^2$ we get $ 4|x$ or $ 4|y$ and $ 3|xy$, n must be $ n=2^{k_1}3^{k_2}p_3^{k_3}..., k_1\ge 2,k_2\ge 1$. Therefore $ d_1=1,d_2=2,d_3=3,d_4=4.$ It is easy to show, that number of prime divisors $ m\ge 3$. If $ 5|n$, then $ d_5=5,d_6=6,d_7\le 10$. If $ d_7=7$ we get $ d_7=4^2-3^2,d_{10}=2*4*3=24,d_{k+1-22}=25=d_{11}$ give contradition with $ 32=k=(k_1+1)(k_2+1)(k_3+1)(k_4+1)...\ge 36$. If $ d_7=8$, then $ d_{10}=15,d_{k+1-22}=17$ give solution \[ n=2^3*3*5*17.\] If $ d_7=9$, then $ d_{10}=40,d_{k+1-22}=41=d_{11}$ or $ d_{10}=12,d_{k+1-22}=15$. In first case $ k\ge 48$ -contradition, For second case $ m=3$ and $ n=2^{k_1}3^{k_2}5^{k_3},(k_1+1)(k_2+1)(k_3+1)=32,k_1\ge 2,k_2\ge 2 \to n=2^3*3^3*5$ contradition $ d_7=8\not =9$. If $ d_7=10$ we get $ d_{10}=24,d_{k+1-22}=26=d_{11}\to n=2^3*3*5*13$ - contradition with $ d_7=8\not =10$. We consider all cases, when $ 5|n$. Let $ 5\not |n$, then $ d_5=6$. If $ 7|n$ then $ d_6=7,d_7\le 12$. ${ d_7=10\to d_{10}=15}$ give contradition, $ d_7=9, d_{10}=40$ give contradition too, $ d_7=11\to d_{10}=60,d_{k+1-22}=61=d_{11}$ give contradition too. If $ d_7=12$, then ${ d_{10}=16,d_{k+1-22}=20}$ give contradition too. Let $ (35,n)=1$. If $ 8\not |n$, then $ 3|k$. If $ 9|n$, then $ d_6=9$ and $ d_7=11,d_{10}=60,d_{11}=61$ or $ d_7=12,d_{1k+1-22}=20$ give contradition. If $ n=4*3*p_3^{k_3}...$, then $ p_3=11$ give $ p_7=12$ - contradition. $ p_3>12$ give $ d_6=12,d_7=p_3,d_{10}=\frac{p_3^2-1}{2},d_{11}=d_{10}+1$ - contradition. Therefore, if $ (35,n)=1$, then $ 8|n$ and $ d_6=8,d_7\le 12$ give not solutions. Therefore we get unique solution $ n=2^3*3*5*17$.

See also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=21999