Let 16P+1=x^3, for some P, a prime, and x, a whole number.
Then (x-1) (x^2+x+1) =16P.
Say (x^2 + x + 1)=0,mod 16.
But that is impossible, since (x^2 + x + 1) is always odd, so never divisible by 16.
Therefore (x-1)=0,mod 16, so let x= 16n+1.
Now 16P+1=(16n+1)^3, so that
16P=(16n)*((16n+1)^2+(16n+1)+1)
Note the first part 16P=(16n)..., making n a factor of P
So P can be prime iff n=1.
((16+1)^2+(16+1)+1) =307, a prime, and we are done.
Edited on July 13, 2020, 10:51 pm
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Posted by broll
on 2020-07-13 22:45:58 |
Possible solution
