4
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There is a similar case like this already on the site, but it deals with perfect squares and is relatively easy to solve.

But what about perfect cubes? Thus $3p+1= n^3$ ?

any help? Thanks!

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    Something else to consider: $n^3 \equiv 0, 1, 8 \pmod 9$, which means $n \equiv 1 \pmod 3$ for $p$ in $3p + 1 = n^3$ to have a shot at being an integer. Furthermore, $n$ must be even. These are necessary but not sufficient conditions for $p$ to be prime. – 2017-02-17

2 Answers 2

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Why you don't try ?$$n^3-1=3p \to (n-1)(n^2+n+1)=3p \\\to \begin{cases}n-1=p &n^2+n+1=3\\ n-1=3 &n^2+n+1=p \\n-1=3p &n^2+n+1=1\\ n-1=1 &n^2+n+1=3p\end{cases}$$ all of happening cases

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    That was very useful Khosrotash. I can work through it now. Thanks – 2017-02-17
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    I tried all four cases, but I didn't get a solution – 2017-02-17
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    That's it. You got all possible solutions. The number of solutions is 0 and you found them all. – 2017-02-17
  • 1
    There is no solution for all cases – 2017-02-17
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This is not the solution, I tried all four cases and didn't get anywhere.... sigh. Can someone help?

Case 1: assume p=n-1 and 3=(n^2)+n+1 Then n=p+1, so
3=(p+1)^2 + (p+1) +1

3= p^2 +2p +1 +p +1 +1

3= p^2 +3p +3

but then what?

0=p^2 +3p

0= p(p +3) so p is 0 or -3? would 3 be the answer? i thought it would have to be positive...

Case 2: assume 3=n-1 and p=(n^2)+n+1 Then, n= 4

p=16 +4 +1 = 21, which is not prime.

Case 3: assume 3p=n-1 and 1=(n^2)+n+1 Then, n= 3p+1

1= (3p+1)^2 +(3p+1) +1

1= 9p^2 +6p+ 1 +3p+1 +1

1= 9p^2 +9p +3

-2= 9(p^2-p)

-2/9 = p^2-p

which ... not sure about a solution

Case 4:

assume 1=n-1 and 3p=(n^2)+n+1 Then, n=2

3p = 4+2+1

3p = 7

p=7/3 which is not prime nor an integer....

Any help????

  • 1
    Well, you got your solution then. You asked "which primes are there such that 3p+1 is a cube". You found all possible cases, examined them, no solution. So that's it. There is no prime p such that 3p+1 is a cube. Problem solved. – 2017-02-17