jmc

The given equation expands to $$(a^2+b^2)x^2-(4ab+1)x+a^2+b^2=0.$$Since the quadratic has an integer root, its discriminant is nonnegative: $$(4ab+1{)}^2-4(a^2+b^2{)}^2\ge 0.$$This factors as $$(4ab+1+2a^2+2b^2)(4ab+1-2a^2-2b^2)\ge 0.$$We can write this as $$[1+2(a+b{)}^2][1-2(a-b{)}^2]\ge 0.$$Since $1+2(a+b{)}^2$ is always nonnegative, $$1-2(a-b{)}^2\ge 0,$$so $(a-b{)}^2\le \frac{1}{2}.$

Recall that $a$ and $b$ are integers. If $a$ and $b$ are distinct, then $(a-b{)}^2\ge 1,$ so we must have $a=b.$ Then the given equation becomes $$2a^2{x}^2-(4a^2+1)x+2a^2=0.$$Let $r$ and $s$ be the roots, where $r$ is the integer. Then by Vieta's formulas, $$r+s=\frac{4a^2+1}{2a^2}=2+\frac{1}{2a^2},$$and $rs=1.$

Since $rs=1,$ either both $r$ and $s$ are positive, or both $r$ and $s$ are negative. Since $r+s$ is positive, $r$ and $s$ are positive. Since $a$ is an integer, $$r+s=2+\frac{1}{2a^2}<3,$$so the integer $r$ must be 1 or 2. If $r=1,$ then $s=1,$ so both roots are integers, contradiction. Hence, $r=2,$ and $s=\boxed{\frac{1}{2}}.$ (For these values, we can take $a=1.$)