Find all possible values of $ d_{17}$ knowing that $ (d_7)^2 + (d_{15})^2= (d_{16})^2$

Question

Let $ d_1,d_2 \ldots, d_r$ be the positive divisors of $ n$. Namely, $$ 1=d_1<d_2< \ldots <d_r=n$$ Now if $ (d_7)^2 + (d_{15})^2= (d_{16})^2$ find all possible values of $ d_{17}$

I don't know which tools should I use here. I know so far that all the prime divisors satisfy the condition $d^2\le \frac{n}2$

Patently this tells us that must be always some relationships between divisors of a number $n$ satisfying Pythagorian relation such as $ (d_7)^2 + (d_{15})^2= (d_{16})^2$.

Any hint or help is welcome

Answer 1

Hint:

It is well-known, that if $a^2+b^2=c^2$ then $60|abc$ so $d_2=2,d_3=3,d_4=4,d_5=5,d_6=6$ and $d_7=7,8,9$ or $10$

Further Hint

If $d_7=7$ then $d_{16}-d_{15}=1$, hence $d_{16} + d_{15}=49 \implies d_{16}=25,d_{15}=24$ then $4200|n$ and $d_8=8$ and...

$10,12,14,15,20,21,24|n$ \to $d_{15} \leq 24$ but it means ,that $d_{9},d_{10},...,d_{15}=10,12,14,15,20,21,24$ and so $9,13 \not | n$ so $d_{17}=28$