21first few steps:
(a+1)3- a3 = n2
3a2+ 3a+1 = n2
try some factoring (may be both sides) and co-prime , perfect square sort of argument.
If difference of cubes of two consecutive natural numbers is equal to n2 (n is a positive integer). Prove that 2n-1 is a perfect square.
this is not very difficult
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4 Answers
1(a+1)2 + a2 + a(a+1) = n2 ..
where a is a natural number.. dunno what to do from here... :( any hints?
2121after post 3
3a2+ 3a+1 = n2
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.