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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

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

$$2p^{3}=(k+√206)(k-√206)$$

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?

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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".2012-07-23
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    $$p^{3} = 1(mod 9) or 8(mod9)$$ 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 ?2012-07-23
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    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

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