Does it satisfy

$\Huge{7744=88^2}$ Let the the expression be $\text{In decimal form be ......}$ $\Large{xxyy=(kl)^2}$

$\large{\Rightarrow kl \times kl = x0y0 + x0y}$

$\large{\Rightarrow kl \times k0 +kl \times l = x0y0 + x0y}$

$\large{\Rightarrow kl \times k0 = x0y0 \quad \text{and} \quad kl \times l = x0y}$

$\large{\Rightarrow k = l }$

Now,

$\Large{xxyy=(kk)^2}$

$\large{\Rightarrow kk \times k = x0y}$

$\large{\Rightarrow 11k^2 = x0y}$

$\large{\Rightarrow k^2 = \Large{\frac{x0y}{11}}}$

By divisibility rule of 11

$\large{x + y = 11}$

$\Large{\begin{Bmatrix} 7,4& \not4,\not7 \\ 5,6& 6,5\\ \not8,\not3& \not3,\not8\\ 2,9& \not9,\not2 \end{Bmatrix}}$

By trial and error $\large{x = 7 ; y = 4; k = 8 \quad\text{which are also found to be unique}}$

$\Large{\mathfrak{Q.E.D}}$

I know this is a crude way of proving it but this was the first solution that came to my mind. i hope the problem creator posts a better way of doing the same.