Conjecture

[davieddy]

Originally Posted by xilman

Possibly because a Diophantine equation requires solutions over Z and x is not an element of that ring?

Just how many smartasses can one forum stand?

[wblipp]

Originally Posted by devarajkandadai

Is x = l 1 l the only solution to x^2 + 8 = 3^n ?

Devaraj

Yes.

For n even, n=2m, we have 8 = (3^m-x)*(3^m+x). Both factors are integers, so the first factor must be one of -8, -4, -2, -1, 1, 2, 4, 8, and the second factor must be 8/(3^m-x). Checking the eight cases shows there are only the two known solutions for n even.

3^n is odd, so x must be odd. For n odd, we have

(2y+1)^2+9 = 3*9^m

4y^2+4y+9 = 3*9^m

4*(y^2+y) = 3*9^m-9

The left side is always zero mod 4. The right side is always 2 mod 4. Hence there are no solutions for n odd.

[LaurV]

Originally Posted by wblipp

Yes.