The number of integral values of $\alpha$  for which the equation  $x^{26}+(k^2+2k\alpha -\alpha -3)x^6+x^4+3=0$ has no real solution for all real values of k, is

$x^{26}+\left(k^2+2k\alpha -3\right)x^6+x^4+3=0(x=0\text{ can never be a solution })$

$\text{ For }x\ne 0\text{ we have }$

$\begin{array}{l}{k}^2+2k\alpha -\alpha -3=-\left(x^{20}+x^{-2}+3x^{-6}\right)=-\underset{\text{A.theas }25}{\underbrace{-\left(x^{20}+x^{-2}+x^{-6}+x^{-6}+x^{-6}\right)}}\\ \left\{\begin{array}{c}\because A.M\ge G.H.\\ \frac{x^{20}+x^{-2}+x^{-6}+x^{-6}+x^{-6}}{5}\ge {\left(x^{20-2-6-6-6}\right)}^{1/5}\end{array}\right\}\\ \end{array}$\

$$\begin{gathered} \therefore \text{ R.H.S. is always }\le -5 \\ \text{ For this equation to have no solution. } \\ \text{ L.H.S. should be }>-5\mathrm{\forall }k\text{. } \end{gathered}$$

$\begin{array}{l}\therefore k^2+2k\alpha -\alpha -3>-5\mathrm{\forall }k.\\ \Rightarrow k^2+2k\alpha -\alpha +2>0\mathrm{\forall }k.\\ \Rightarrow (2\alpha {)}^2-4.1(-\alpha +2)<0\\ \Rightarrow {\alpha }^2+\alpha -2<0\\ \Rightarrow (\alpha +2)(\alpha -1)<0\\ \therefore \alpha =0\text{ or }-1\end{array}$