jmc

The square of the binomial $rx+s$ is $$(rx+s{)}^2=r^2{x}^2+2rsx+s^2.$$If this is equal to $a{x}^2+15x+4$, then $s$ must be either 2 or -2. Since $(rx+s{)}^2=(-rx-s{)}^2$, we may choose either $s=2$ or $s=-2$, and the solution will be the same. We choose $s=2$.

The square of $rx+2$ is $$(rx+2{)}^2=r^2{x}^2+4rx+4.$$If this is equal to $a{x}^2+15x+4$ then we must have $15=4r$ or $r=\frac{15}{4}$. This gives our square: $${\left(\frac{15}{4}x+2\right)}^2=\frac{225}{16}{x}^2+15x+4.$$Therefore $a=\boxed{\frac{225}{16}}$.