3
$\begingroup$

How many primes $p$ are there such that $2p^{3} + 206$ is a perfect square?

My approach: Let the square be $k^{2}$, then

$2p^{3} + 206 = k^{2}$ $2p^{3}=k^{2} - 206$

$2p^{3}=(k+√206)(k-√206)$

Now, let $(k+√206)=p^{3}; (k-√206)=2$ and

$(k-√206)=p^{3}; (k+√206)=2$

But I couldn't solve it further. Are there no such primes? Am I on the right track? Please help.

I got $19$ by hit-and-trial. Is there any analytic way?

  • 0
    If $2p^3+206$ is a perfect square $k^2$, then $2p^3=k^2-206$ Thus $2\mid k^2$, hence $2\mid k$. Let's say $k=2r$. Then $2p^3=(2r)^2-206=4r^2-206$ and therefore $p^3=2r^2-103$ Modulo 9, we get $p^3\equiv 2r^2+5\bmod 9$ $(0,1,\,\text{or}\;8)\equiv 2\cdot (0,1,4,\,\text{or}\;7)+5\bmod 9$ $(0,1,\,\text{or}\;8)\equiv (0,2,8,\,\text{or}\;5)+5\bmod 9$ $(0,1,\,\text{or}\;8)\equiv (5,7,4,\,\text{or}\;1)\bmod 9$ However, this does not rule out the existence of solutions, because $1$ is a possible value of both sides.2012-07-23

1 Answers 1

11

By a trivial change of variables this is a Mordell equation $y^2 = x^3 + k$ with $k=824$. Therefore there are only finitely many integer solutions, but often it takes effort to bound the sizes of such solutions. For $|k| \le 10000$, all solutions were tabulated by an algorithm of Gebel, Pethő and Zimmer in this paper.

According to the data at OEIS, the equation $y^2 = x^3 + 824$ has exactly two integer solutions, so these would be the $(38, \pm 236)$ that you found already (corresponding to $p=19$). Apparently there are no more solutions, whether or not $p$ is prime.