
#1
Oct804, 10:52 PM

P: 16

Hi, I need help solving this problem. The question asks me to find all prime numbers p such that p^2 = n^3 + 1 for some integer n.




#2
Oct904, 07:14 AM

Sci Advisor
HW Helper
P: 1,123

This is really quite easy as p^2 can only be p times p or 1 times p^2.
Now, [tex]n^3 + 1 = (n + 1)(n^2  n + 1)[/tex] So one possible solution is where one of the factors is equal to 1 and the other is equal to p^2. Or when they are both equal to each other. Test those out and you should find all the possible solutions for n. 



#3
Oct1304, 03:48 PM

P: 646

An alternative,
all odd p^2 is usually of the form p^2 = 1 mod 4 n^3+1 = 1 mod 4 or n^3 = 0 mod 4 Now this is useful if and only if u have a list of primes ... So first u can generate a list and then check for conditions ...  AI 


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