The affine point $(0,1)$ lies on the elliptic curve $E:y^2=x^3+x+1$ over $\mathbb Q$. Find an affine point on $E$ with rational, non-integral coordinates.
Well I have found with the tangent method: $(1/4,9/8)$ is on the curve, but what if the point had integer coordinates ?
So my question is: Why does the task require to find a non-integral point ? does it emphasize the impossibility of such case here, or is there a method to obtain rational, non-integral coordinates from integer coordinates ?
Ok, so I suppose you added the point $\;(0,1)\;$ to itself (as it is a "vertex" point on the curve) by the tangent method, and you found the above...but you could also take any rational line through $\;(0,1)\;$ and take some intersection point with the curve, for example:
$$\ell: y=\frac12x+1\implies\frac14x^2+x+1=x^3+x+1\implies x^3-\frac14x^2=0\iff$$
$$x^2\left(x-\frac14\right)=0\implies x=0,\,x=\frac14\implies y=1,\,y=\frac98$$
and then you get a new point on the curve: $\;\left(\cfrac14,\,\cfrac98\right)\;$ (or two, with $\;\pm\frac98\;$) .
With any rational choice of the slope above you get new points, though it can be those poinbts on the curve won't have, in general, rational coordinates. For this wwe would have to use a straight line through two rational points, so that the third root will also have to be rational as well.
For example, we already have the rational points $\;(0,1),\,\left(\frac14,\,-\frac98\right)\;$ , and the straight line through them is
$$m:=\frac{-\frac{17}8}{\frac14}=-\frac{17}2\implies y-1=-\frac{17}2x\implies y=-\frac{17}2x+1$$
and substituting in our cubic:
$$\frac{289}4x^2-17x+1=x^3+x+1\implies x\left(4x^2-289x+72\right)=0$$
The quadratic's discriminant is:
$$289^2-4\cdot288=287^2\implies x_{1,2}=\frac{289\pm287}8=\begin{cases}72\\{}\\\cfrac14\end{cases}$$
and we get two new rational points: $\;\left(72,\,\pm611\right)\;$