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Find all primes $p$ such that $2p^2-3p-1$ is a perfect cube.

I've noticed that $p=2,3$ are solutions. Setting $2p^2-3p-1=x^3$ we can factorise this as $(4p-3)^2=8x^3+17$. However im unable to proceed further.

1 Answers 1

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Let $$2p^2-3p-1=x^3$$ $$p(2p-3)=(x+1)(x^2-x+1)$$ If $p=2$ then $x=1$

If $p=3$ then $2p^2-3p-1=8=2^3$

Let $p>3$. Then $\gcd(p,2p-3)=1$ and $p<2p-3$

Then $$p=x+1$$ $$2p-3=x^2-x+1$$

Thus $$-3=x^2-3x-1$$ $$x^2-3x+2=0$$ Hence $x=1$ or $x=2$

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    Your claim that $p > 3$ implies $p = x + 1$ is not justified. It's conceivable that $p$ is a proper factor of $x^2-x+1$. – 2017-01-08
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    Let $$2p^2-3p-1=m^3,$$ then: $$p(2p-3)=(m+1)(m^2-m+1).$$ Let $m>4$. Then: 1) $p|m^2-m+1$, 2) $p>m+1$, 3) $m+1|2p-3$ ( $2p=3\pmod{m+1}$). $m^2-m+1=3\pmod{m+1}$, then $m^2-m+1=2p\pmod{m+1},$ $m^2-m+1=2p+q(m)(m+1)$ for some $q(m)\in\mathbb{N}$. $p|q(m)$, Then: $$m^2-m+1=p(2+q(m)(m+1)).$$ Let $q(m)\ge 1$, then: $m^2-m+1\ge p(m+3)\ge(m+2)(m+3)$, contradition. Let $q(m)=0$ ($m^2-m+1=2p$),(in this case $p(2p-3)=(m+1)(m^2-m+1)=2p(m+1)$, ($2p-3=2(m+1)$, $2|3$). – 2017-01-08
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    Ok, but then the argument in your original answer very much incomplete. SInce p is prime, the equality $p(2p-3) = x^3+1$ decomposes into two cases: (1) $p$ divides $x +1$; (2) $p$ divides $x^2 - x + 1$. Each case needs to be handled separately. Note that in your original argument, you didn't address case (2) at all, and as you can see, case (2) is the harder of the two cases. – 2017-01-08
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    @Roman8_3 -- In your latest comment you use $q(m)$ both for the original $q(m)$ and for $q(m)/p$. They need separate names. – 2017-01-09
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    @Roman8_3 -- I think you should edit your answer to show the complete analysis. As originally posted, it has unacceptable gaps. – 2017-01-09
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    @quasi: Ok! I will do it – 2017-01-09