Squares of the digits

Let us call two digits number $A= 10x+ y$ where $x$ and $y$ are it's digits. As per conditions supplied $\begin{aligned} & x^2 +y^2 = A-1 \\& x^2+ y^2 = 10x+y -1 \\& x^2 -10x +y^2 -y +1 =0 \\& x = \frac{10\pm\sqrt{100 - 4(y^2-y+1)}}{2} \\& x = 5\pm\sqrt{25-(y^2-y+1)} \\& x =5\pm\sqrt{25-p}\end{aligned}$ Here $x$ can only be integer iff $\sqrt{25-p }$ is a perfect square number within the range of $9\leq p\leq 21$ . Therefore, $x$ has an integer solutions for $p = 21, 16 , 9$ however, the integer solution of $y= 5 , -4$ is only defined for $p= y^2-y+1 =21$ . So as a whole we can claim that the integer solutions for $(x,y)$ exist at $p=21$ which are $(7,5), (3,5)$ .

Thus required largest two digits number is $\boxed{75}$ .