Given:
x2−8x+32ax2+2(a+1)x+9a+4<0∀x∈R
For quadratic x2−8x+32=0,D1=(−8)2−4(32)=−64
Since the discriminant is less than zero and the leading coefficient is positive, this quadratic will always be positive.
Now, solving ax2+2(a+1)x+9a+4<0
We know that, for a quadratic to be always negative, the coefficient of x2<0,D<0.
⇒a<0
But we want positive values.
So, no positive integral value exist.