**·**2011-08-03 00:18:42

after post 3

3a^{2}+ 3a+1 = n^{2}

3(2a+1)^{2}= (2n+1)(2n-1)

gcd(2n+1,2n-1) = 1

**case 1: 3|2n+1,** then (2a+1)^{2} = (2n+1)3(2n-1) , gcd( (2n+1)3,(2n-1))= 1

of course (2n-1) is a perfect square.

**case 2: 3|2n-1** , we will show 3 can't divide 2n-1. Cz if 3|2n-1, 2n-1 = 3(2k+1) for some integer k. or 2n+1 = 6k+5. Again

(2a+1)^{2} = (2n-1)3(2n+1) , gcd( (2n-1)3,(2n+1))= 1

of course (2n+1) is a perfect square. But it is of the form 6k+5. No square can be of this form. Hence we are done.

PS. In the sheet it was proved by Pell equations. Pls tell me if something looks wrong in the previous proof.