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

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

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


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?

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You can't just "let" $$(k+\sqrt{206})=p^3;(k-\sqrt{206})=2,$$ both in the sense of "there is no reason to believe that the factorization needs to be of that form, so to assume it is true might throw away some solutions" and in the sense of "that is impossible because $k$, $p^3$, and $2$ are integers, while $\sqrt{206}$ is not". – Zev Chonoles Jul 23 '12 at 7:56
$$p^{3} = 1(mod 9) or 8(mod9)$$ and perfect square only have digit sum as $1,4,9, or 7$ $$2*8 + 8 = 6 Mod 9$$ so not possible $$2*1 + 8 = 1 Mod 9$$ say $$2*(9m+1)+8 = k^{2}$$ or $$m = (k^{2} -10)/18$$ have no integral solutions. So I guess no such prime exits ? – Bazinga Jul 23 '12 at 8:06
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. – Zev Chonoles Jul 23 '12 at 8:13

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.

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