Given homogeneous function $p$ with order $m$, how can I show that
$$p^{-1}(a) = \left(\frac{a}{b}\right)^{\frac{1}{m}}p^{-1}(b)?$$
The original question is:
Let $p$ be any homogeneous polynomial in $k$-variables. Homogeneity means $$p(tx_1, \dots, tx_k) = t^mp(x_1, \dots, x_k).$$ Prove that the set of points $x$, where $p(x) = a$, is a $k-1$ dimensional submanifold of $\mathbb{R}^k$, provided that $a \neq 0$. Show that the manifolds obtained with $a > 0$ are all diffeomorphic, as are those with $a <0$.
So I am wondering, is $m$ is odd, $a,b$ does not need to have the same sign to be homegeneous, since $\lambda = \left(\frac{b}{a}\right)^{1/m}$ still make sense? Thanks.
According to @Daniel Fischer's kind advice:
$$\vec{v} \in p^{-1}(a)$$ $$\lambda \vec{v} \in p^{-1}(\lambda^m a) \Rightarrow \vec{v} \in \frac{1}{\lambda} p^{-1}(\lambda^m a)$$ So assuming the preimage is unique, we have $$p^{-1}(a) = \frac{1}{\lambda} p^{-1}(\lambda^m a)$$ Set $\lambda = \big(\frac{b}{a}\big)^{\frac{1}{m}}$, we get $$p^{-1}(a) = \Big(\frac{b}{a}\Big)^{-\frac{1}{m}} p^{-1}\Big(\Big(\frac{b}{a}\Big)^{\frac{1}{m} \cdot m}a\Big) \Rightarrow p^{-1}(a) = \Big(\frac{a}{b}\Big)^\frac{1}{m} p^{-1}(a)$$
Though, I'm not certain about assuming uniqueness.