It is nothing but a problem with real quadratic forms. You have a pair of vectors $v,v' \in R^4$ with, respectively, components $(\Delta t, \Delta x, \Delta y, \Delta z)$ and $(\Delta t', \Delta x', \Delta y', \Delta z')$. Actually these components describe the same vector in spacetime (describing the difference of events) but referring to two different reference frames.
Next you consider two quadratic forms $$ds^2(v,v)=
c^2\Delta t^2 - \Delta x^2 - \Delta y^2 -\Delta z^2$$ and $$ds'^2(v',v')=
c^2\Delta t'^2 - \Delta x'^2 - \Delta y'^2 -\Delta z'^2\:.$$
Since there is a linear transformation connecting the two reference frames (which will be proven to be the Lorentz transformation!) we also have $v'= Av$, where $A$ is some non-singular $4 \times 4$ real matrix independent from $v$, we can also write:
$$ds'^2(v',v')= ds'^2(Av,Av)$$
We finally know, from physics, that:
$$ds'^2(v',v') =0 \quad \mbox{if and only if}\quad ds^2(v,v) =0\:,$$
that is
$$ds'^2(Av,Av) =0 \quad \mbox{if and only if}\quad ds^2(v,v) =0\:.$$
In other words:
the quadratic forms $ds^2(\:,\:)$ and $ds'^2(A\cdot\:,A\cdot)$ have the same zeros.
A theorem of quadratic forms states that:
THEORM. Two real quadratic forms on a real $n$ dimensional vector space, respectively associated to the symmetric matrix $$\eta = diag(1,-1, \ldots, -1)$$
and to the symmetric matrix $\eta'$, have the same zeros if and only if they are proportional, i.e.,
$$\eta' = c \eta \quad \mbox{for some $c\in \mathbb R \setminus\{0\}$}\:.$$
Applying the result to our case we have that there is $a\neq 0$ with
$$ds^2(v,v)= ads'^2(Av,Av)\quad \mbox{for every $v\in R^4$}.$$
I do not have Landau's book, so I do not know how it is subsequently proved therein that $a=1$. I think that some physical symmetry argument is exploited in addition to the invariance of $c$.
ADDED REMARKS
1.
PROOF of the Theorem.
Due to the form of $\eta$, we have that $U^t\eta U =0$ if and only if $U_V^0 = \pm |V|$ and $U_V^{i}= V_{i}$ for $V \in \mathbb R^{n-1}$ and $i=1,\ldots,n-1$, where $|V| = \sqrt{V_1^2+\cdots+V_{n-1}^2}$. Our hypotheses on $\eta'$ can therefore be made explicit as follows.
$$U_V^t\eta'U_V=0 \quad \forall V \in \mathbb R^{n-1}\:.$$
In components (latin indices are summed from $1$ to $n-1$)
$$\eta'_{00}|V|^2 \pm \sum_i \eta'_{0i} |V| V_i + \sum_{ij} \eta'_{ij} V_iV_j=0\:.$$
We henceforth use the sign $+$, the other case can be treated similarly. It must be
$$\eta'_{00}|V|^2 + \sum_i \eta'_{0i} |V| V_i + \sum_{ij} \eta'_{ij} V_iV_j=0\quad \forall V \in \mathbb R^{n-1}\:.\tag{1}$$
Since (1) holds for $V$ and $-V$, we immediately have that
$$\sum_i \eta'_{0i} |V| V_i=0 \quad \forall V \in \mathbb R^{n-1}\:.$$
Arbitrariness of $V$ easily implies that $\eta'_{0i}=0$ for all $i=1,\ldots, n-1$.
(1) can be re-written as follows
$$\sum_{ij} (\eta'_{00}\delta_{ij}+\eta'_{ij}) V_iV_j=0\quad \forall V \in \mathbb R^{n-1}\:.\tag{2}$$
Since $\eta'_{ij}=\eta'_{ji}$, using $V=X+Y$ and next $V=X-Y$, (2) yields
$$\sum_{ij} (\eta'_{00}\delta_{ij}+\eta'_{ij}) X_iY_j=0\quad \forall X,Y \in \mathbb R^{n-1}\:.\tag{3}$$
In other words the matrix inside the scalar product above is the zero matrix,
$$\eta'_{00}\delta_{ij}+\eta'_{ij}=0\:. $$
Summing up
$$\eta'= \eta'_{00} diag (1, -1, \cdots, -1)\:.\tag{4}$$
Notice that $\eta'_{00}$ cannot vanish otherwise the zeros of $\eta'$ would fill the whole vector space, differently form the zeros of $\eta$, but we know that the set of zeros must coincide by hypotheses. (4) is the thesis if $c:= \eta'_{00}$. QED
2. Evidently, making use of Sylvester's theorem, the statement of the proved theorem implies an apparently stronger form of it like this.
THEORM. Consider two real quadratic forms on a real $n$ dimensional vector space, respectively associated to the symmetric matrices $\eta$ and $\eta'$, where $\eta$ has signature $+,-,\ldots, -$.
$\eta$ and $\eta'$ have the same zeros if and only if they are proportional
