I would like to solve if it is possible next diophantine equation for positive integers $n\geq 1$, $x\geq 1$ and $y\geq 1$

$$n x^2-\operatorname{rad}(n)=y^3,\tag{1}$$ where $\operatorname{rad}(n)$ denotes the product of distinc primes dividing our integer $n>1$ with the definition $\operatorname{rad}(1)=1$ (it is the arithmetic function Radical of an integer from Wikipedia).

Computational facts. Using a Pari/GP program the solutions, we write these as $(x,y,n)$, that I can to find in next segments $1\leq x\leq 500$, $1\leq y\leq 500$ and $1\leq n\leq 500$ are respectively $(x,y,n)=(3,2,1)$, $(17,12,6), (26,15,5),(99,70,35)$ and $(485,198,33)$. Thus I've no enough computational evidence to say if these are a complete set of solutions pf $(1)$.

Claim 0 (Preparation). For each fixed $n>1$ (see Claim 1) and being $\prod_{p\mid n}p^{e_p}$ the prime factorization of $n$ (fundamental theorem of arithmetic) we are searching/we need to characterize integers with the form $$\left(\prod_{p\mid n}p^{e_p-1}\right)\cdot x^2-1\tag{2}$$ and that are divisible by $p^2$ for each prime $p$ dividing $n$.

Question. I would like to know a full characterization of the solutions of our equation, or at least to know if there exist infinitely many or finitely many of such solutions. I wrote previous Claim 0 that I believe that is the key result that combined to Euler-Fermat theorem will provide us an inventory of solutions. I add also auxiliary claims that I think that will be useful to finish this task. Can you find the solutions of $$n x^2-\operatorname{rad}(n)=y^3,$$ over positive integers? Is there a finite number of such solutions $(x,y,n)$? Many thanks.

Claim 1. The case $n=1$ is Mihăilescu's theorem (see the for example the Wikipedia's article dedicated to Catalan's conjecture).

Claim 2. It's easy to prove that $n$ has no repeated prime factors.

Proof. By contradiction let $p\mid n$ such that $p^2\mid n$ then from $$n\cdot x^2-y^3=\operatorname{rad}(n)$$ we deduce the absurd that $p^2\mid \operatorname{rad}(n).$ We've used the obvious fact that for each prime $p$ dividing $n$ then $p^3\mid y^3$ holds$\square$.

Claim 3. One has that for each solution $(x,y,n)$ the condition $\gcd(x,n)=1$ is satisfied.

Proof. For the case $n=1$ we invoke again Mihăilescu's theorem. By contradiction, for $n>1$ we assume that there exists a prime $p\mid \gcd(x,n)$ and we get a contradiction using a similar argument than previous proof. $\square$

  • $\begingroup$ I add here the calculation of the particular values of the Euler's totient function for each $n$ in one of those solutions that I've shown, these are $\varphi(n)=1,2,4,24$ and $20$, respectively for $1, 6, 5, 35$ and $33$. Also I add that I can deduce a statement for $(1)$ on assumption of the abc conjecture (I didn't add it, because maybe isn't useful). $\endgroup$ – user243301 Jun 2 '18 at 14:24
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    $\begingroup$ The theory of elliptic curves might be helpful. $\endgroup$ – Peter Jun 3 '18 at 8:34
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    $\begingroup$ A further solution is $$[3363, 2378, 1189]$$ $\endgroup$ – Peter Jun 3 '18 at 8:47
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    $\begingroup$ You proved that $n$ must be squarefree. So, you can simplify your equation to $$n\cdot x^2-n=y^3$$ $\endgroup$ – Peter Jun 3 '18 at 8:57
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    $\begingroup$ I just found out that the solutions of the pell-equation $$a^2-2b^2=1$$ lead to a solution of $(1)$ whenever $\frac{b}{2}$ is a squarefree integer. Probably, this gives infinite many solutions. $\endgroup$ – Peter Jun 3 '18 at 9:14

Partial result :

We want to show that the equation $$nx^2-n=y^3$$ has infinite many integer solutions with positive squarefree $n$ and positive integers $x,y$

The given equation is equivalent to $$x^2-\frac{y}{n}\cdot y^2=1$$

Assume $x,y$ is a solution of the pell-equation $$x^2-2y^2=1$$ and $\frac{y}{2}$ is a positive squarefree integer. Then with $n:=\frac{y}{2}$, we get the desired solution. Hence, $(1)$ has probably infinite many solutions.

The first solutions of this kind are :

? for(k=1,45,w=[3,4;2,3]^k*[3;2];n=w[2,1];if(issquarefree(n/2)==1,print(w)))
[17; 12]
[99; 70]
[3363; 2378]
[3880899; 2744210]
[22619537; 15994428]
[131836323; 93222358]
[4478554083; 3166815962]
[5168247530883; 3654502875938]
[175568277047523; 124145519261542]
[34761632124320657; 24580185800219268]
[202605639573839043; 143263821649299118]
[6882627592338442563; 4866752642924153522]
[40114893348711941777; 28365513113449345692]
[233806732499933208099; 165326326037771920630]
[7942546277405390632803; 5616228332641321147898]
[269812766699283348307203; 190786436983767147107902]
[9165691521498228451812099; 6481122629115441680520770]
[53421565080956452077519377; 37774750930342781945186508]
[10577200073262678228000529443; 7479209897770887057999820682]
[61648439810871582916000871057; 43592029839838017630000520932]
[359313438791966819268004696899; 254072969141257218722003304910]
[12206079718853609176884159165123; 8631001740904974549490112546258]
[414647397002230745194793406917283; 293199986221627877463941823267862]
[82098090374248746619236402542311697; 58052116426097312032565778987728148]
  • $\begingroup$ Many thanks I'm going to study it, and to read about Pell's equation. $\endgroup$ – user243301 Jun 3 '18 at 9:21
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    $\begingroup$ We can use other values for $\frac{y}{n}$, for example $19$, giving this solution $$[626611046296874412523432743802509517466276016770, 14375443303759799795917894808 7846788095935677961, 7566022791452526208377839373044567794522930419]$$ $\endgroup$ – Peter Jun 3 '18 at 15:22
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    $\begingroup$ Many thanks for you effort, if this equation $n x^2-\operatorname{rad}(n)=y^3$ isn't in the literature I would like it to be called Peter-Pell equation. $\endgroup$ – user243301 Jun 3 '18 at 17:10

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