Sum of two squares

[anoy9021]

If the difference of the cubes of two consecutive number is square number,then the root of that square number can be written the sum of squares of 2 consecutive numbers.
Example: $8^3-7^3=13^2,13=2^2+3^2$

[*Mahi*]

Why have you posted the same thing twice? There is no need to do this.

[Masum]

If the difference of the cubes of two consecutive number is square number,then the root of that square number can be written the sum of squares of 2 consecutive numbers.
Example: $8^3-7^3=13^2,13=2^2+3^2$

Firstly, see the reasons of editing. And don't make double posts, it creates confusion. I have deleted the second one. In these cases, contact with the moderators.
Anyway, here is a solution.
$\text{Solution}$ :
$\ \ \ \ (x+1)^3-x^3=y^2$
$\Longrightarrow y^2=3x^2+3x+1$
$\Longrightarrow 4y^2=3(4x^2+4x+1)+1$
$\Longrightarrow (2y+1)(2y-1)=3(2x+1)^2$
Note that $gcd(2y+1,2y-1)=1$.
So let $(2x+1)=mn$ with $gcd(m,n)=1$
We have two cases :
$\#1$. $2y+1=m^2,2y-1=3n^2$.
This implies $m^2-3n^2=2\Longrightarrow m^2\equiv-1\pmod3$, contradiction!
$\#2$. $2y+1=3m^2,2y-1=n^2$.
Then $n$ odd, say $n=2t+1$, we have $y=t^2+(t+1)^2$,as desired.

[*Mahi*]

I didn't , honestly!