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Is there any result on solutions of diophantine equation $n^3+3=m^2$?

I only know solution $\left(1, 2\right)$ and I checked that there are no other solution up to 100000. But I can't prove that generally.

And is there anything more general on equation $n^p+p=m^2$? I can solve it for $p = 2$ and $p = 7$.

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  • $\begingroup$ @lulu : Are you able to find the "Your Answer" box below, in which your answer belongs? $\endgroup$ Commented Jul 16, 2017 at 20:57
  • $\begingroup$ @lulu , what I think Eric Towers meant to say was, "That's a very helpful comment. Why not post it as an answer, which OP can then accept?" $\endgroup$ Commented Jul 16, 2017 at 21:03
  • $\begingroup$ Note: I have deleted my prior comment, and the posted solution. A number of closely related problems have been resolved, but it is not immediately clear from a search of the literature whether this particular case has been settled. $\endgroup$
    – lulu
    Commented Jul 16, 2017 at 21:45

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This can also be written as $m^2=n^3+3$ which is an elliptic curve. They are usually written as $y^2 = x^3 + Ax+B$ where $(x,y)$ are the variables instead of $(m,n)$. This particular type where $A=0$ are called Mordell equations. These are well studied, you could look for "integer points on elliptic curves", see for example following answer for a quite good list on known bounds and algorithms:

It is known (Siegel theorem) that there are at most finitely many integer solutions.

One method of finding new rational points on the curve is "adding" the known points (in the group of the curve). In our case we only known one point. By "doubling" this known point $(1,2)$ we get a non-integral point $(-23/16,-11/64)$, so Lutz Nagell tells us that we cannot find further integral points with this method.

Using the bound mentioned in this answer and the mentioned CAS, we find that there are indeed no further integral points. (Other than the obvious $(1,-2)$)


Regarding your other question: for $p=2$ we get $m^2-n^2 = 2$ which is a hyperbola. You can easily determine all solutions by observing $2 = m^2-n^2 = (m-n)(m+n)$.

I don't think that any other $p$ have been studied, other than perhaps $p=5$ as a hyperelliptic curve.

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With quadratic and cubic mod 4 "values" you can show that

$$n^3 \equiv_4 1 \implies n \equiv _41 \iff \exists k \in \mathbb{N}: n=4k+1$$

$$m^2 \equiv_4 0 \implies m \equiv _40 \iff \exists h \in \mathbb{N}: m=2h$$

in fact

$$n^3-1 \equiv_4 n^3+3=m^2$$

Is easy to see that $m^2 \equiv_4 0, 1$ and $n^3 \equiv_40,1,-1$.
RHS is positive so $n^3 \equiv_41$. $\blacksquare$

These two condiction reduces the cases, you have to find $(n,m)$ with the form $(4k+1, 2h)$.

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