= Solution
Insert an exact solution into the <linear multistep method> and use $f(t,y(t))=y'(t)$. Its local residual is
$$
R_h(t)=\sum_{\ell=0}^s\rho_\ell y(t+\ell h)-h\sum_{\ell=0}^s\sigma_\ell y'(t+\ell h).
$$
Expanding both terms by <Taylor theorem> gives
$$
R_h(t)=C_0y(t)+\sum_{k=1}^p\frac{h^k}{k!}C_k y^{(k)}(t)+O(h^{p+1}),\qquad C_0=\sum_\ell\rho_\ell,\quad C_k=\sum_\ell\ell^k\rho_\ell-k\sum_\ell\ell^{k-1}\sigma_\ell.
$$
Consequently $C_0=\cdots=C_p=0$ is sufficient for order at least $p$. It is also necessary: test the residual on the polynomials $y(t)=1,t,\ldots,t^p$, evaluating at $t=0$. For $y(t)=t^k$, it equals $h^k C_k$ (and equals $C_0$ for the constant), which can be $O(h^{p+1})$ only if $C_k=0$. These functions can be realized as exact solutions of suitable smooth <ordinary differential equations>, so this is a genuine necessity test. Thus \b[the order conditions are]
$$
\boxed{\sum_{\ell=0}^s\rho_\ell=0,\qquad \sum_{\ell=0}^s\ell^k\rho_\ell=k\sum_{\ell=0}^s\ell^{k-1}\sigma_\ell\quad(1\leq k\leq p).}
$$
This is a statement about local order; <zero-stability> is the separate condition that turns it into convergent approximation.
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