Question

for any positive integer p, we have cube root of (-p) = ____​

Answer 1

Answer:

3

Step-by-step explanation:

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Answer 2

Answer:

Suppose 16p+1=n3. This rearranges to

16p=(n−1)(n2+n+1).

Since n2+n+1 is an odd factor of 16p that is greater than 1 , p is odd and p=n2+n+1.

This forces n−1=16, so n=17 is the only possibility. Indeed this gives p=307, which is indeed prime and hence the only solution.

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