Let $x = \lfloor \sqrt{2n} \rfloor$
Hence $x^2 <= 2n < (x+1)^2$
Or $x^2 <= 2n < x^2+2x +1$
Or $2n = x ^2 + p$ where $0 <= p < 2x +1\cdots(1)$
Now from the given condition
$1+ x | x^2 + p$ as $2n = x^2 + p\cdots(2)$
Or $1 + x | (x^1-1) + (p+1)$
And as $1+ x | (x^2-1)$
We have $ 1 + x | p+1$
x cannot be zero as x=0 gives n = 0 which is not natural number
Let $p+1 = m(1+x)$
As $p + 1 < 2x+2$
So $m(1+x) < 2x +2$ from (1)
Or $m < 2$
And m is not -ve
As we have $m=0$ or $m = 1$
m=0 gives p = -1 which is not admissible
m= 1 gives $p = x$ or $n= \frac{x(x+1)}{2}$
so n is number of the form $n= \frac{x(x+1)}{2}$
