SOLUTION: If n is a positive integer such that 2n+1 is a perfect square, show that n+1 is the sum of two successive perfect squares

Since 2n + 1 is odd, then since 2n + 1 is a perfect square, 
its square root is also odd, say 2p + 1, where p > 0, then

 2n + 1 = (2p + 1)²
 2n + 1 = 4p² + 4p + 1
     2n = 4p² + 4p
      n = 2p² + 2p
      n = p² + p² + 2p
  n + 1 = p² + p² + 2p + 1
  n + 1 = p² + (p + 1)²

Edwin