The nonnegative integer solutions for the following equations:
x 3 + 1 = y 2 {\displaystyle x^3+1=y^2}
x ( x + 1 ) 2 = y ( y + 1 ) ( y + 2 ) 6 {\displaystyle \frac{x(x+1)}{2}=\frac{y(y+1)(y+2)}{6}}
x 2 = y ( y + 1 ) ( y + 2 ) 6 {\displaystyle x^2=\frac{y(y+1)(y+2)}{6}}
x ( x + 1 ) 2 = y ( y + 1 ) ( 2 y + 1 ) 6 {\displaystyle \frac{x(x+1)}{2}=\frac{y(y+1)(2y+1)}{6}}
x 2 = y ( y + 1 ) ( 2 y + 1 ) 6 {\displaystyle x^2=\frac{y(y+1)(2y+1)}{6}}
2 x − 7 = y 2 {\displaystyle 2^x-7=y^2}
x ! + 1 = y 2 {\displaystyle x!+1=y^2}
x # = y ( y + 1 ) {\displaystyle x\#=y(y+1)}
x 2 − n y 2 = 1 {\displaystyle x^2-ny^2=1}
