jmc

We have that $y^2+24y+16+k=(y+b{)}^2=y^2+2by+b^2$ for some $b.$ Since $2by=24y,$ we see that $b=12.$ Now, expanding $(y+12{)}^2$ gives us $y^2+24x+144,$ so $16+k=144,$ so $k=\boxed{128}.$