(In reply to
Explanation to Puzzle Answer by K Sengupta)
At the outset, we observe that, when p=2 then 16p+1=33, which is NOT a perfect cube.
Now, 16p+1=r^3 (say)
=> 16p= r^3-1
=> 16p =(r-1)(r^2+r+1)
=> p ={(r-1)/16}*(r^2+r+1)
CASE 1: r is even
Then, r= 2n, for some integer n
p = odd/even
So, p cannot be an integer.
Contradiction.
CASE 2: r is odd
<=> r= 2n+1 for some integer n.
Then, p = (even/even)* odd = even/even, which is a valid candidate for further exploration.
Then,
2n+1-1 2n n
p = ------------- * (r^2+r+1) = ------ *(r^2+r+1) = ------*(r^2+r+1)
16 16 8
If (n,r)= (1,3) , then: p =13/8, which is not an integer. Contradiction.
If (n,r) = (8,17), then: p = 17^2+17+1=307
Now, floor(V307) =17
Dividing 307 with all odd numbers <17, we observe that the quotient in each case is NOT an integer.
Consequently, p=307 is the smallest prime number satisfying the given conditions.
Explanation to Puzzle Answer: Method II
