By Mahler's classical theorem on continuous $p$-adic functions, $p$-pod sequences are precisely those that extend to continuous functions on $\mathbb{Z}_p$. So the problem comes down to: the product of two continuous functions is continuous, which is obvious. Of course, there should also be a direct approach...

The crucial lemma of my solution is that a sequence $z_n$ is $p$-pod iff for each $e \geq 0$ there is a polynomial $P_e$ such that $z_n \equiv P_e(n)$ mod $p^e$ for all $n$.

Sung-yoon Kim wrote: The crucial lemma of my solution is that a sequence $z_n$ is $p$-pod iff for each $e \geq 0$ there is a polynomial $P_e$ such that $z_n \equiv P_e(n)$ mod $p^e$ for all $n$. I'm not sure about my answers My solution in $Z_{p^e}$ Using the Interpolation formular,we find that: $f(x)=\sum_{k=0}^{m}z_k\coprod_{j\neq k}\frac{x-j}{k-j}$ Assume $deg f\leq m-1$,we have $\sum_{k=0}^{m}\frac{(-1)^{m-k}}{k!(m-k)!}z_k=0$ Then $\sum_{k=0}^{m}(-1)^k\binom{m}{k}z_k=0$ Select $m$ enough to large,because a Polynomials have finite root. It's easy to see: A sequence $z_n$ is $p$-pod iff for each $e \geq 0$ there is a polynomial $P_e$ such that $z_n \equiv P_e(n)$ mod $p^e$ for all $n$.(*) From (*) we complete the proof.

jejungchv wrote: Sung-yoon Kim wrote: The crucial lemma of my solution is that a sequence $z_n$ is $p$-pod iff for each $e \geq 0$ there is a polynomial $P_e$ such that $z_n \equiv P_e(n)$ mod $p^e$ for all $n$. I'm not sure about my answers My solution in $Z_{p^e}$ Using the Interpolation formular,we find that: $f(x)=\sum_{k=0}^{m}z_k\coprod_{j\neq k}\frac{x-j}{k-j}$ Assume $deg f\leq m-1$,we have $\sum_{k=0}^{m}\frac{(-1)^{m-k}}{k!(m-k)!}z_k=0$ Then $\sum_{k=0}^{m}(-1)^k\binom{m}{k}z_k=0$ Select $m$ enough to large,because a Polynomials have finite root. It's easy to see: A sequence $z_n$ is $p$-pod iff for each $e \geq 0$ there is a polynomial $P_e$ such that $z_n \equiv P_e(n)$ mod $p^e$ for all $n$.(*) From (*) we complete the proof. why can we assume that deg(f)<m and f(x) seems to be related to m,who can explain it to me

Wait, easy way...

Finite differences. If one knows the basic properties of a finite difference especially the Leibniz rule (how to find the finite difference of a product of two functions), he/she has 99% chance to solve this problem. By definition $\Delta_h f(x):=f(x+h)-f(x)\,,\, \Delta^m_h f(x)=\Delta_h\left(\Delta_h^{m-1} f \right)(x)$. It's called the finite difference of order $m$ with step $h$ taken at a point $x$. It can be written explicitly as: $$\Delta_h^m f(x)=\sum_{j=0}^m (-1)^{m-j}\binom{m}{j} f(x+jh) $$To express the finite difference of a product of two functions $f,g$ using the finite differences of $f$ and $g$ we use the following identity (the Leibniz rule) $$(1)\,\,\,\,\,\,\,\,\,\, \Delta_h^m (f\cdot g)(x)=\sum_{j=0}^m \binom{m}{j}\Delta_h^j f(x)\Delta_h^{m-j} g(x+jh) $$ Back to the OP. To stick to the above notations, let $f(n), g(n), n\in \mathbb{N}$ be the functions/sequences corresponding to $\{x_n\}_{n=0}^{\infty}$ and $\{y_n\}_{n=0}^{\infty}$. We want to show that if $f$ and $g$ are $p$-pod then $fg$ is also $p$-pod. The definition of the $p$-pod just says $\Delta_1^m f(0)$ and $\Delta_1^m g(0)$ are multiple of $p^e$ when $m$ is big enough. Fix some $e\in\mathbb{N}$ and let $m$ be a big number. Look at $(1)$. Clearly $\Delta_1^j f(0)$ is multiple of $p^e$ when $j$ is big enough. It's enough to show that when $j$ is small, $\Delta_1^{m-j} g(j)$ is also multiple of $p^e$. It boils down to show the truncated sequence $g(j), g(j+1),\dots$ is also $p$-pod. It can be shown using the definition of finite difference and induction on $j$. Indeed, $\Delta^m_1 g(j-1)=\Delta_1^{m-1}g(j)-\Delta_1^{m-1} g(j-1)$. Putting consecutively $j=1,2,\dots$ we get that $\Delta_1^{m-1} g(j)$ is multiple of $p^e$ when $m$ is large enough. It completes the proof.

I hope I didn't fakesolve Let $P_m$ be a polynomial of degree at most $m$ such that $P_m(x)=z_x$ for each $0\le x\le m$. Let $P_m(x)=\sum\limits_{i=0}^m b_i \binom{x}{i}$ Then $S_m=\sum\limits_{k=0}^m (-1)^k {m \choose k} P_m(k)=\sum\limits_{k=0}^m (-1)^k {m \choose k} \sum\limits_{i=0}^m b_i\binom ki$ $= \sum\limits_{i=0}^m b_i\binom mi \sum\limits_{k=0}^m (-1)^k \binom{m-i}{m-k}$ Notice $\sum\limits_{k=0}^m (-1)^k \binom{m-i}{m-k} = \sum\limits_{k=i}^m (-1)^k \binom{m-i}{m-k}$. Let $u=m-k$, then this sum is $\sum\limits_{u=0}^{m-i} \binom{m-i}{u} (-1)^{m-u} = (-1)^m \sum\limits_{u=0}^{m-i} \binom{m-i}{u} (-1)^u$. Notice the inner sum vanishes unless $m-i=0$, in which case it computes to $1$. Therefore, this entire sum is simply $b_m(-1)^m$, so we want $p^e\mid b_m$. One observation is that $b_m$ is simply $m!$ multiplied by the leading coefficient of $P$ (note that $b_m\in \mathbb{Z}, $ but $\frac{b_m}{m!}$ is not necessarily an integer.) Consider the extension step: note $P_{m+1}(x)-P_m(x)=0$ for all $0\le x\le m, x\in\mathbb{Z}$, so $P_{m+1}(x)-P_m(x)=cx(x-1)\cdots (x-m)=c\prod\limits_{i=0}^m (x-i)$. We can see $|S_{m+1}|=c(m+1)!=P_{m+1}(m+1) - P_m(m+1) = z_{m+1}-P_m(m+1)$. (On a side note, this is exactly what you get by Lagrange Interpolation anyways) We can now see for all sufficiently large $m,n$, $p^e\mid P_{m+1}(n)-P_m(n)=(z_{m+1}-P_m(m+1))\binom{x}{m+1}$, so $p^e\mid x_n-P_m(n)$. Repeat the same argument for $y_n$ and use the polynomial $Q_k$. Therefore, for all $n$ large, $x_ny_n\equiv P_m(n)Q_m(n) (\bmod\; p^e)$ Note $P_m(t)Q_m(t)=x_ty_t$ for all $0\le t\le m$, so the same properties for $P_m(n)$ also applies to $P_m(n)Q_m(n)$. Since those properties are just a rephrasement of conditions of p-pods, we are done.

Two words: finite differences. For each $e$, the condition is equivalent to the existence of a polynomial $P$ such that $P(n) \equiv z_n \pmod {p^e}$. On the other hand for $P(n), Q(n)$ characteristic polynomials of $\{a_n\}$ and $\{b_n\}$, obviously $$PQ(n) \equiv a_nb_n \pmod {p^e},$$done.

This problem is so dumb... The following claim is an instant kill. Claim: If $\{z_n\}_{n=0}^\infty$ is a $p$-pod sequence, then for each $e\geq 0$, there exists a polynomial $R(t)\in \mathbb{Q}[ t ]$ such that $z_n-R(n)$ is an integer divided by $p^e$ for $n\geq 0$. Proof of the claim: Suppose that $p^e$ divides $\sum_{k=0}^m (-1)^k {m \choose k} z_k$ for $m\geq N$. Set $R(t)$ to be the Lagrange interpolation polynomial on $(0,z_0),\dots,(N-1,z_{N-1})$. $\blacksquare$