I'm trying to solve this problem: $$n!+n(n+1)/2=m^3.$$ If $n+1=p$ is prime this leads to $2(n-1)!+p=a^3.$ If $p$ is a cubic non-residue for some $q<p$ the equation has no solution. This seems to work for $p>13$. I know we only need to consider $3k+1$ primes for $q$ and that a possibility might be to consider the cubic residues of sucessive relevant primes and with CRT draw conclusions on $p$. I find this approach confusing. What i would like is to use a cubic reciprocity argument but i've searched online and can't understand how to use it. It's pretty obvious that after some point a result like this holds, so there must be a simple argument.
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1Hello and welcome to math.stackexchange. This is an interesting problem, thank you for sharing your thoughts on possible approaches. As to your approach - how are $a$ and $m$ related? – Hans Engler Mar 05 '25 at 19:53
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Thank you @HansEngler. From the initial equation we can factor $n/2$ and we get two co-prime factors, both have to be cubes. So $n/2=b^3$ and $n!/(n/2)+n+1=a^3$ with $n+1=p$. From this we have $p=2 b^3+1$. Applying Wilson's theorem to the second equation we also know $2$ has to be a cubic residue $\mod p$ – Fernando Cagarrinho Mar 06 '25 at 06:59
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It's not really that obvious to me that it's true. Heuristically I would expect it to be true for sufficiently large $p$, as naively every prime $q<p$ has probability $\tfrac13$ of $p$ being a nonresidue. So the probability of a counterexample rapidly drops as $(\tfrac13)^{p/\log p}$, expecting only finitely many failures, probably all very small. But a proof seems out of reach. – Servaes Mar 06 '25 at 09:44
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If the problem was about squares, it would be easy to solve. If $p$ is $4n+1$ you just need to find an odd QNR $mod p$ and apply quadratic reciprocity. If $p$ is $4n+3$ you need an odd $4n+1$ QNR or an odd $4n+3$ QR. It's simple to prove that for both forms of $p$ we can find the required residue. In this case i thought that something similar, necessarely more evolved, could be done with cubic reciprocity. My graduation is in physics and i'm interested on this things for about 2 years. With what i could find online i'm unable to use CR efectively. – Fernando Cagarrinho Mar 06 '25 at 15:56
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Thank you @Servaes for your input. – Fernando Cagarrinho Mar 06 '25 at 16:07
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@FernandoCagarrinho, If we use formula $[\frac n2(n+1)]^n=n^n(n!)$ we get $n!=(\frac{n+1}2)^n$ which gives $(\frac{n+1}2)^n+\frac n2(n+1)=m^3$, let $\frac n2(n+1)=S$ then we have: $(\frac S n)^n+S=m^3$.May be concentrating on this equation make the job easier. – sirous Apr 08 '25 at 08:17
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1https://mathoverflow.net/q/492391/520102 – Tong Lingling May 14 '25 at 06:39