Let $p_i$ and $q_i$ be the roots in ascending order. 1) $p_1\le q_1$. $P(x)=p(x+p_1)$ and $Q(x)=q(x+p_1)$ still satisfy the condition. $Q(0)=-1$ implies all roots $q_i-p_1$ equal to $1$. So $Q(x)=(x-1)^n$ with $n$ odd. We need to find $P(x)=(x-1)^n+1$ has $n$ non-negative roots $P_i$. If $n>1$, $\sum P_i=n$ and $\sum P_i^2=n$ imply all $P_i=1$. Absurd! The only solution is $n=1$. $p(x)=x-a$ and $q(x)=x-a-1$ where $a\ge0$. 2) $q_1\le p_1$. $P(x)=p(x+q_1)$ and $Q(x)=q(x+q_1)$ still satisfy the condition. $P(0)=1$ implies all roots $p_i-q_1$ equal to $1$. $P(x)=(x-1)^n$ with $n$ even. Similarly we can show no solution exists for $n>2$ . The other solution is $n=2$. $p(x)=(x-a)^2$ and $q(x)=(x-a-1)(x-a+1)$ where $a\ge1$.

Let $p(x)=(x- \alpha_1)(x-\alpha_2)...(x- \alpha_n)$ and $q(x)=p(x)-1=(x- \beta_1)(x-\beta_2)...(x- \beta_n)$ WLOG assume that they are arranged in increasing order. From the equation they must satisfy, we see that if we plug in $x=\alpha_i$ we have $q(\alpha_i)+1=0$. Since all the roots are integers, we need that every $(\alpha_i- \beta_j)$ has to be either $1$ or $-1$. Obviously, they can't be all 1's, so for every $\alpha_i$ there exists an integer $1 \leq m \leq n$ for which $\alpha_i- \beta_m=-1$ and because the roots are in order, this implies $\alpha_i- \beta_k=-1$ for $k\geq m$, in particular let's take the $m$ from $\alpha_1$ this means that $\beta_m = \beta{m+1}=...=\beta_n (*)$ . Since $\alpha_i- \beta_m \geq \alpha_1 -\beta_m=1$ if the $m$ corresponding to $\alpha_i$ were greater than $\alpha_1$'s $m$ it would contradict $(*)$ thus; \[ \alpha_1=\alpha_2=...=\alpha_n= \beta_1+1 = \beta_m -1 := \alpha \]Then the equation takes the form \[ (x-\alpha)^n=(x-\alpha+1)^{m-1}(x-\alpha-1)^{n-m+1}+1 \]And we need that $n-m+1$ to be odd. Plugging $x=\alpha-1$; $(-1)^n = 1$ for this to happen $n$ must be even, and thus so must $m$. Next do $x=\alpha-2$ then; \[ (-2)^n = (-1)^{m-1}(-3)^{n-m+1} \implies 2^n=3^{n-m+1} +1 \]Similarly if $x=\alpha +2$ \[ (2)^n = (3)^{m-1}(1)^{n-m+1} \implies 2^n=3^{m-1} +1 \]Thus $3^{m-1}=3^{n-m+1} $ meaning $m-1=n-m+1 \implies n=2m-2$ substituting in the original equation we get; \[((x-\alpha)^2)^{m-1} -((x-\alpha)^2-1)^{m-1}=1 \]This is possible iff $m-1=1$ and hence $n=2$ this gives us the solution \[ p(x)=(x-\alpha)^2, q(x)= (x-\alpha)^2-1\]Where $\alpha$ is an integer greater or equal than one. There exists the possibility, however, that $m=1$ In which case the equation forces $n$ to be odd and furthermore $n=1$ getting the solutions \[ p(x)= x- \alpha , q(x)= x- \alpha -1\]Where $\alpha \geq 0$

first of all $p(x)=1+q(x)$ set $p(x)=(x-\alpha_{1})(x-\alpha_{2})(x-\alpha_{3})\cdots (x-\alpha_{n})$ $q(x)=(x-\beta_{1})(x-\beta_{2})(x-\beta_{3})\cdots (x-\beta_{n})$ s.t. $\alpha_{1}\le \alpha_{2}\le \alpha_{3} \cdots \le \alpha_{n}$ and $\beta_{1}\le \beta_{2} \le \beta_{3}\cdots \le \beta_{n}$ where $\alpha_{i} , \beta_{i} \in \mathbb{Z^{*}}$ case1:- $\alpha_{n}\ge \beta_{i}$ $\forall$ $1\le i\le n$ Claim:- only $n=1$ works pf:-$q(\alpha_{1})=\prod_{i=1}^{n} (\alpha_{1}-\beta_{i})$ clearly $q(\alpha_{n})=-1 \implies \prod_{i=1}^{n} (\alpha_{n}-\beta_{i})=-1 \implies \beta_{1}=\beta_{2}=\beta_{3}=\cdots =\beta_{n}:=\beta \in \mathbb{Z^{*}}$ so $q(x)=(x-\beta)^{n}$ and $p(x)=(x-\beta)^{n}+1\implies n$ has to be odd now $\sum_{i\neq j} \alpha_{i}\cdot \alpha_{j}= \frac{n(n-1)\beta^2}{2}, \sum {\alpha_{i}^2}=n\beta^2$ from C-S we get: $\left(\sum_{i\neq j} \alpha_{i} \cdot \alpha_{j} \right)^2 \le \left (\sum \alpha_{i}^2 \right)^2 \implies (n-1)^2\le 4\implies n\le 3$ so clearly $n=1$ or $3$ but clearly $n=3$ don't works as $(x-\beta)^3+1$ has not all roots to be integer so we get $n=1$ to be working , hence claim follows $\blacksquare$ $p(x)=x-\beta+1$ and $q(x)=x-\beta$ Case 2:- $\beta_{n} \ge \alpha_{i}$ $\forall$ $1\le i \le n \in \mathbb{Z^{*}}$ Claim:- only $n=2$ works pf:-clearly $p(\beta_{n})=\prod_{i=1}^{n} (\beta_{n}-\alpha_{i})$ so $p(\beta_{n})=q(\beta_{n})+1\implies p(\beta_{n})=1 \implies \prod_{i=1}^{n} (\beta_{n}-\alpha_{i})=1 \implies \alpha_{1}=\alpha_{2}=\alpha_{3}=\cdots \alpha_{n}:=\alpha \in \mathbb{Z^{*}}$ so $p(x)=(x-\alpha)^{n}$ and $q(x)=(x-\alpha)^{n}-1\implies n$ has to be even now again $\sum_{i\neq j} \beta_{i}\cdot \beta_{j}=\frac{n(n-1)\alpha^2}{2}$ and $\sum \beta_{i}^2=n\alpha^2$ again from C-S we get: $n\le 3$ , so $n=2$ works as claim follows $\blacksquare$ hence $p(x)=(x-\alpha)^2$ and $q(x)=(x-\alpha)^2-1$ so clubbing all solutions we got we get: $\boxed{(p(x)=x-\beta+1, q(x)=x-\beta ), (p(x)=(x-\alpha)^2 , q(x)=(x-\alpha)^2-1)}$ to be the polynomials satisfying given condition for some $\alpha , \beta \in \mathbb{Z^{*}}$ $\blacksquare$