Binomial coefficients: a minimum value problem

The $x^4$ term is $\dbinom{5}{4}a(bx)^4=5ab^4x^4$ .

The $x^2$ term is $\dbinom{5}{2}a^3(bx)^2=10a^3b^2x^2$ .

We then have that $\begin{aligned}5ab^4&=8(10a^3b^2)\\\iff 5ab^4&=80a^3b^2\\\iff 5b^2&=80a^2\because a,b\neq 0\\\iff b^2&=16a^2\\\iff (4a-b)(4a+b)&=0\end{aligned}$

Since $a,b>0$ we have $4a\neq -b$ , so $4a=b$ , and $a+b=5a$ . The smallest value of this for positive integers $a$ occurs when $a=1$ , so the answer is $\color{#20A900}{\boxed{5}}$ .