Projective coordinates for Montgomery curves

I have this Montgomery curve $y^2=x^3+10x^2+x \mod 83$ and a point $Q(3,28)$, doubling this point in affine coordinates I get $2Q(61,35)$.

Switching to projective coordinates I know that $x = X/Z$ and $y = Y/Z$ and $Z = 1$ at the beginning so $Q(3,28,1)$, following the Wikipedia formulas for doubling in projective coordinates, I get $X_2p = 64$ and $Z_2p = 65$.

How do I verify that this point in projective coordinates belongs to the curve? Do I need the $y$-coordinate in projective coordinates? If so, how do I calculate it?

How do I verify that this point in projective coordinates belongs to the curve?

Assuming that you don't have the $y$ coordinate, then the requirement on $x$ is that $x^3 + 10x^2 + x$ is a quadratic residue; that is, there exists a $y$ such that $y^2$ is that value.

Now, you have the projective coordinates; the corresponding requirement is that:

$$(Y/Z)^2 = (X/Z)^3 + 10(X/Z)^2 + (X/Z)$$

Multiplying both sides by $Z^4$, we get

$$(YZ)^2 = X^3Z + 10X^2Z^2 + XZ^3$$

The lhs is a square (and by selecting the correct $Y$, can be any square, assuming $Z \ne 0$), and so the condition is that $X^3Z + 10X^2Z^2 + XZ^3$ is a quadratic residue.

I believe that testing this value for quadratic residuosity will be cheaper than recovering the possible $y$ values.