What are the prime numbers that satisfy k²=n³+1 when n is not a prime number?

[TrifidBlue]

Homework Statement

If k is a prime number
find all k that satisfy k²=n³+1
n is not a prime number

Homework Equations

I really have no idea, use any suitable one

The Attempt at a Solution

all prime numbers are odd except 2.
n must be positive natural number
n³ = k² -1 = (k-1)(k+1)

any help is appreciated

 

[Tusike]

Two things you need to know to solve this problem:
1) If k is prime, then k^2 can only be divided by 4 numbers, 2 of which are k^2 and 1.
2) a^3 + b^3 = (a^2-ab+b^2)(a+b)

If you need more clues, ask, but post anything you did with these clues:)

-Tusike

 

[TrifidBlue]

Two things you need to know to solve this problem:
1) If k is prime, then k^2 can only be divided by 4 numbers, 2 of which are k^2 and 1.
2) a^3 + b^3 = (a^2-ab+b^2)(a+b)

If you need more clues, ask, but post anything you did with these clues:)

-Tusike

thanks I'll try these hints

 

[TrifidBlue]

I tried with these two clues,
k^2 = (n+1)(n^2 -n+1)
k = sqrt [(n+1)(n^2 +n+1)]
and from clue (1) k^2 can be divided by k^2 , 1, k and -k
but I don't know how to connect the information together to get the answer :(

 

[Tusike]

Ugh sorry, I said something wrong
k^2 can only be divided by 3 numbers, not 4 (I was thinking how it can be written down two ways as a multiple of two numbers...) . So what does that say about (n+1) and (n^2 -n + 1)?

 

[TrifidBlue]

Ugh sorry, I said something wrong
k^2 can only be divided by 3 numbers, not 4 (I was thinking how it can be written down two ways as a multiple of two numbers...) . So what does that say about (n+1) and (n^2 -n + 1)?

k^2= (n+1) [ n(n-1) +1]
that is
k^2 = n (n+1)(n-1) + (n+1)
right?

 

[Tusike]

You don't need to solve that equation just yet, only see how it can happen.
On the left side, you have k^2, and on the right side you have a multiple of two whole numbers, (n+1) and (n^2 - n + 1).
You know that k^2 can either be written as 1*k^2, k^2*1, or sqr(k^2)*sqr(k^2). These are also multiples of two whole numbers! So just pair them up:) You should get n=2, and from then k=3. 1*k^2 and k^2 * 1 won't lead to anything worth noticing.

EDIT: sqr(k^2), not sqr(k), sorry...

 

[TrifidBlue]

You don't need to solve that equation just yet, only see how it can happen.
On the left side, you have k^2, and on the right side you have a multiple of two whole numbers, (n+1) and (n^2 - n + 1).
You know that k^2 can either be written as 1*k^2, k^2*1, or sqr(k^2)*sqr(k^2). These are also multiples of two whole numbers! So just pair them up:) You should get n=2, and from then k=3. 1*k^2 and k^2 * 1 won't lead to anything worth noticing.

EDIT: sqr(k^2), not sqr(k), sorry...

thaaaaaaaaaaaaaank you!
please check the steps
(n+1) = k
(n^2 -n +1) = k
n^2 -n +1 = n+1
n(n-2)=0 n=0 , k=1 (not prime)
n=2 , k=3

 

[Tusike]

Yes, that is correct. Now I don't know about how "precise" this proof has to be, but if has to be really precise, you might want to examine the following two possibilities as well:
(1)
(n+1) = 1 and (n^2 - n + 1) = k^2

(2)
(n+1) = k^2 and (n^2 - n + 1) = 1

From the first you'll get that n = 0 and k = 1, which of course isn't prime, so that's not a good solution.
From the second, you'll get that n = 1, and k=sqr(2), which again isn't prime. However, you couldn't of known that you won't get valid solutions from these two, so I think they need to be added to the proof as well to make it complete.

 

[TrifidBlue]

Thank you very much,Tusike
^_^