<s>[INST] What values for a prime $p$ are possible if $3p+1$ is a perfect cube? 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!
 [/INST] 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????
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