Is there a positive integer $n$ such that $n=d_6 ^ 2 + d_7 ^ 2 -1$, where $d_k$ is $k^{th}$ divisor of $n$? ( $1$ is first, etc., $n$ is last) So far have only concluded that $n$ in that case should have more than $12$ divisors, since $n=d_6 ^ 2 + d_7 ^ 2 -1 > d_6 d_7$ and $d_6$and $d_7$ are not in the middle of array of divisors of $n$. Also, if $d_6$ and $d_7$ are relatively prime, then we would need to solve equation $d_6 d_7 m = d_6 ^ 2 + d_7 ^ 2 -1$, but they are not necessarily relatively prime. Thanks for reading, any help is appreciated. (problem came from one friend)
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6Does this answer your question? An identity satisfying the divisors of a positive integer - found using an Approach0 search. – John Omielan Dec 24 '22 at 01:42
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3If $\gcd(d_6,d_7)=m>1$ then, since $m,|,n$ we'd get $m,|,1$, a contradiction. – lulu Dec 24 '22 at 01:42
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1"(problem came from one friend)" You can tell this one friend, that he/she can use a search for the stackexchange site. – Dietrich Burde Dec 26 '22 at 09:33