I think that it can be proved that a polinomial with coeff in $\{\pm 1\}$ is irreducible. try to do this

You're wrong - try to prove it yourself

I don't remember very well this fact.. I'll try a proof: Assume that $Q=P(x^2+ax+b)$ have the desired form. Since $Q(0)=bP(0)$ meand that $b=\pm 1$. Assume that $b=1$. The other case is similar. We will prove that $a\le 2$. Assume that $a\ge 3$. Denote by $b_i$ coeff of $P$. We will have : $b_m=1$ $ab_m+b_{m-1}=c_{n-1}=\pm 1$ .... $b_{i-1}+ab_i+b_{i+1}=c_{i-2}= \pm 1$ ... Take $b_i$ the maximum of $b_j,j=1,..m$ in absolute value. We will have (There are more cases but this one is the worse): $(a-2)|b_i|+|b_{i-1}|+|b_{i+1}|\le a|b_i|= |-b_{i-1} \pm 1-b_{i+1}|\le |b_{i-1}|+|b_{i+1}|+1$. Hence $|b_i|\le 1$. Moreover, if $a\ge 4$ the conclusion follows imediatly from the above argument, because we will get $|b_i|\le \frac 1 2$. If $a=3$ we have that $3b_m+b_{m-1}=\pm 1\Rightarrow b_{m-1}=3\pm 1\ge 1$. Contradiction. Hence the solutions are: $(1,\pm 1);(-1,\pm 1);(2,\pm 1);(-2,\pm 1);(0,\pm 1)$, and the verification is simple. I hope that I discussed all cases.

I have a simpler proof, I think. First note that if $R(x)=x^n+c_{n-1}x^{n-1}+\ldots+c_0$ where $c_i\in \{-1,1\}$ and $a\in \mathbb C$ and $|a|\geq 2$ then $R(a)\neq 0$. It is simply proved using the triangle inequality: \[ |R(a)|=|a^n+c_{n-1}a^{n-1}+\ldots+c_0|\geq |a|^n-(|c_{n-1}a^{n-1}|+|c_{n-2}a^{n-2}|+\ldots+|c_0|)=|a|^n-(|a|^{n-1}+|a|^{n-2}+\ldots+|a|^0)=|a|^n-\frac{|a|^n-1}{|a|-1}\geq |a|^n-(|a|^n-1)=1 \] Next see that $b=1$ or $b=-1$. Now using standard and simple methods (like seeing the expression of roots of $x^2+ax+b$) we derive that $x^2+ax+b$ has both roots with norm less than $2$ and $b=1$ or $b=-1$ if and only if \[ (a,b)\in \{(-2,1),(-1,1),(0,1),(1,1),(2,1),(-1,-1),(0,-1),(1,-1)\} \] So the only possible solutions are from this set. But all of them are answers (you can check that very easily, just for each case take $P(x)=1$ or $P(x)=x+1$ or $P(x)=x-1$)

xirti wrote: Hence the solutions are: $(1,\pm 1);(-1,\pm 1);(2,\pm 1);(-2,\pm 1);(0,\pm 1)$, and the verification is simple. I hope that I discussed all cases. The cases $x^2+2x-1$ and $x^2-2x-1$ are not solutions (for example according to my proof).

Actually I didn't do the verification. These cases can be checked manually.

It also appeared in our prepration exam 7 days ago. So I think everyone knows where it comes from!

Yeah, that's clear by now

To Nima Ahmadi Pour: Can you post more of the problems from your preparation exam?

My method: Consider the coefficient $a_{0},...,a_{n-2}$ of $P(x)$ and take the $i$ such that $|a_{i}|$ is largest now use $|c_{i}|=1$ to prove that $|a|\le 3$ and that it can be $3$ only when $|a_{i}|\le 1$ for all $i$. But this leads easily to a contradiction(when considering the first 2 coefficients or whatever) giving $|a|\le 2$. Now rule out the polynomials $|a|=2$ and $b=-1$ by induction on coeficients of $P(x)$(like just use $|c_{i}|=1$ incuctively on $i$ eith from $n$ or from $0$ depending on what polynomial you are considering.

It is easy to see that $b \in [-1.1]$.Also if $|a| \ge 2$ then note the roots of the quadratic are real.If $\alpha,\beta$ be the roots then from the fact that $c_i \in [-1,1]$ we get that $|\alpha|,|\beta| <2$.Thus $|a|=|\alpha+\beta| \le |\alpha|+|\beta|<4$.Thus since $a$ is an integer we get $|a| \le 3$,and thus a bit of case checking gives the solutions $(a,b)=(\pm1,\pm1),(0,\pm1),(\pm2,1)$

Troll...

Can anyone here kind enough point out where I went wrong?

$ P(x)=a_{n-2}x^{n-2} +a_{n-3}x^{n-3} + ... +a_0$ Notice that $b=\pm 1$ , if $|a|>2$ we prove by induction that $|a_{i+1}|>|a_i|$ but we see that $a_{n-2}=1$ so $P(x)=1$ but still still dont work, hence $|a|\leq 2$ and we find the solutions mentioned above

Solution with Aatman Supkar, Aditya Khurmi, Alan, Anant Mudgal, Arindam, Derek Liu, Ethan Liu, Jeffrey Kwan, Eric Shen (CAN), Leo Zipeng Lin, Maximus Lu, Misheel Otgonbayar, Paul Hamrick, R Correaa, Robin, Rohan Goyal, Sean Li: The answer is the following: $a = \pm 1$ and $b = \pm 1$, in this case take $P(x) \equiv 1$; $a = 0$ and $b = \pm 1$, in this case take $P(x) \equiv x+1$; $a = 2$ and $b = 1$; in this case take $P(x) \equiv x-1$. $a = -2$ and $b = 1$; in this case take $P(x) \equiv x+1$. We now show these are the only pairs. The main idea is: Claim: Any root $z$ of $x^2+ax+b$ satisfies $|z| \le 2$. Proof. Since $z$ is a root of $(x^2+ax+b) \cdot P(x)$, we have \[ |z|^n = \left\lvert \sum_0^{n-1} c_{i} z^{i} \right\rvert \le \sum_0^{n-1} \left\lvert z \right\rvert^i \]with the last inequality be triangle inequality. So if $|z| \ge 2$, then run into a contradiction. $\blacksquare$ This claim is enough to win. Indeed, by considering the constant coefficient, we must have $b = \pm 1$. Thus if $|a| \ge 3$, then the roots of $x^2+ax\pm1$ are \[ \frac{|a| + \sqrt{|a|^2 \pm 4}}{2} \]and one of these is greater than $2$. The same is true for $|a|=2$ and $b=-1$. Thus there are no answers other than the ones we claimed.

We claim the solutions are $(a, b) = (c, 1), (d, -1)$ where $c, d$ are integers such that $-2 \leq c \leq 2$ and $-1 \leq d \leq 1$. Main Lemma: If $z$ is a complex root of $x^2 + ax + b$, $|z| < 2$. Proof: By triangle inequality, \begin{align*} |z^n| &= \left|\sum_{i = 0}^{n - 1} c_iz^i\right| \\ |z|^n &\leq \sum_{i = 0}^{n - 1} |z|^i \\ &= \frac{|z|^n - 1}{|z| - 1}. \end{align*}If $|z| \geq 2$, we have \begin{align*} |z|^{n + 1} - |z|^n &\leq |z|^n - 1 \\ |z|^{n + 1} - 2|z|^n + 1 &\leq 0 \\ |z|^n(|z| - 2) + 1 &\leq 0, \end{align*}which is false, contradiction. $\blacksquare$ Note that $bP(0) = c_0 = \pm 1$. As $b$ and $P(0)$ are integers, it follows that $b = \pm 1$. Now, suppose $b = 1$. If $a \geq 3$, we have $\frac{-a - \sqrt{a^2 - 4}}{2} \leq \frac{-3 - \sqrt{5}}{2} < -2$, which is impossible by the lemma. If $a \leq -3$, we have $\frac{-a + \sqrt{a^2 - 4}}{2} \geq \frac{3 + \sqrt{5}}{2} > 2$. Constructions for $-2 \leq a \leq 2$ are demonstrated below: \begin{align*} (x^2 - 2x + 1)(x + 1) &= x^3 - x^2 - x + 1 \\ (x^2 - x + 1)(1) &= x^2 - x + 1 \\ (x^2 + 1)(x + 1) &= x^3 + x^2 + x + 1 \\ (x^2 + x + 1)(1) &= x^2 + x + 1 \\ (x^2 + 2x + 1)(x - 1) = x^3 + x^2 - x - 1. \end{align*} Suppose $b = -1$. If $a \geq 2$, we have $\frac{-2 - \sqrt{a^2 + 4}}{2} \leq \frac{-2 - \sqrt{8}}{2} < -2$, which is impossible by the lemma. If $a \leq -2$, we have $\frac{-a + \sqrt{a^2 + 4}}{2} \geq \frac{2 + \sqrt{8}}{2} > 2$, which is impossible by the lemma. Constructions for $-1 \leq a \leq 1$ are demonstrated below: \begin{align*} (x^2 - x - 1)(1) &= x^2 - x - 1 \\ (x^2 - 1)(x + 1) &= x^3 + x^2 - x - 1 \\ (x^2 + x - 1)(1) &= x^2 + x - 1. \end{align*} Thus, the solutions claimed initially are the only ones, completing the proof. $\Box$

Cute problem. The only $(a, b)$ that work are $(a, b) = (\pm 1, \pm 1)$, $(a,b) = (\pm2, 1), (a,b) = (0, \pm1)$. This clearly works by setting $P(x) = 1$, and the RHS as $x^{2} + ax + b$ in the first case, setting $P(x) = (x+1)$ or $(x-1)$ for the second case, and setting $P(x) = (x+1)$ for the third case. Let $f$ be the RHS. Clearly $b = \pm 1$, because $b | f(0)$ (since $P(x) \in \mathbb{Z}[x]$). Observe that any root $z$ of $f$ must satisfy $\frac{1}{2} < |z| < 2$. If $|z|\leq \frac{1}{2}$, then \[|x^{n} + c_{n-1}x^{n-1} + \ldots c_{1}x| \leq |x^{n}| + |x^{n-1}| + \ldots |x| < 1\](by the geometric sequence formula), a contradiction. We can similarly show $|z| < 2$. Now, we have two cases: Case 1: $b = 1$. Then, the root of $x^{2} + ax + 1$ are $\frac{-a \pm \sqrt{a^{2} - 4}}{2}$. If $a > 2$, then $|\frac{-a - \sqrt{a^{2} - 4}}{2}| > |\frac{-2a+1}{2}| > 2$, a contradiction. Case 2: $b = -1$. Then, the root of $x^{2} + ax - 1$ are $\frac{-a\pm \sqrt{a^{2} + 4}}{2}$. If $a\geq 2$, then $|\frac{-a-\sqrt{a^{2} + 4}}{2}| > |\frac{-2a}{2}| \geq 2$, again a contradiction. Therefore, our only solutions can be $(a, b) = (\pm 1, \pm 1), (\pm 2, 1) (0, \pm 1)$

The answer is $(a,b)=(\pm 1, \pm 1),(0,\pm 1),(\pm 2,1)$. We can take $P(x)=1,P(x)=x+1,P(x)=x-1$ for these cases respectively. We now show that these are the only pairs that work. Here are two solutions: Solution 1 (murders the problem): It is known (proved by Cauchy) that the magnitude of any root of the polynomial $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ is at most: $$1+\text{max}\left\{\left|\frac{a_{n-1}}{a_n}\right|,\left|\frac{a_{n-2}}{a_n}\right|,\ldots,\left|\frac{a_{0}}{a_n}\right|\right\},$$which implies that the roots of $(x^2+ax+b)P(x)$ have magnitude at most $2$. Also note that we require $b=\pm 1$, otherwise $c_0 \neq 1$. If $b=1$, then the roots are $\frac{-a\pm \sqrt{a^2-4}}{2}$. We can see that if $|a| \geq 3$ then we have $\left|\frac{-a-\sqrt{a^2-4}}{2}\right|>2.$ If $b=-1$, then the roots are $\frac{-a\pm \sqrt{a^2+4}}{2}$. We can see that if $|a| \geq 2$ then we have $\left|\frac{-a\pm \sqrt{a^2+4}}{2}\right|>2.$ This resolves the problem. $\blacksquare$ Solution 2: Does not murder the problem: As before note that $b$ must equal $\pm 1$. Further, note that the constant coefficient of $P$ must be $\pm 1$ for the same reason. Also note that we can arbitrarily pick a sign for $a$, since after replacing $a$ with $-a$ we can also replace $x$ with $-x$ (and replace the new $P$ with $-P$ if $n$ is odd), which preserves all the requirements. Denote $P(x)=a_{n-2}x^{n-2}+a_{n-3}x^{n-3}+\cdots+a_1x+a_0$. Note that $a_{n-2}=1$ We consider the following cases: Case 1: $b=1$. We can see that we require $a_{n-3}=-a\pm 1$. We also note that by equating coefficients, for all $0 \leq k \leq n-4$, we require: $$a_{k+2}+a\cdot a_{k+1}+a_k=\pm 1 \implies a_k=-a\cdot a_{k+1}-a_{k+2}\pm 1.$$As mentioned before, we can WLOG assume that $a$ is negative. We now claim that for $|a|>2$, the sequence $a_{n-2},a_{n-3},\ldots,a_1,a_0$ is increasing (strictly). We will prove this with induction. The first two terms are increasing since $a<-2$, which is our base case. Now assume that the sequence $a_{n-2},a_{n-3},\ldots,a_{k+1}$ is increasing for some $k\leq n-4$. Since $a_{n-2}=1$ this also implies that all these terms are positive. We then have: \begin{align*} a_k&=-a\cdot a_{k+1}-a_{k+2}\pm 1\\ &> -(a+1)a_{k+1}-1\\ &\geq 2a_{k+1}-1\\ &\geq a_{k+1} \end{align*}This completes the induction, which implies that $a_0>a_{n-2}=1$. But we must have $a_0 \in \{-1,1\}$, so this is a contradiction. Thus $|a|>2$ with $b=1$ is impossible. Case 2: $b=-1$. We get $a_{n-3}=-a \pm 1$, and the relation: $$-a_{k+2}+a\cdot a_{k+1}+a_k=\pm 1 \implies a_k=a_{k+2}-a\cdot a_{k+1}\pm 1.$$Also WLOG $a$ negative. We claim that for $|a|>1$, the sequence $a_{n-3},\ldots,a_1,a_0$ is increasing, and we also have $a_{n-2} \leq a_{n-3}$. We will prove this with induction. First we note that $a_{n-2}=1 \leq a_{n-3}=-a\pm 1$ since $|a|>1$. Now suppose that $a_{n-2},a_{n-3},a_{n-4},\ldots, a_{k+1}$ is nondecreasing, for some $k \leq n-4$ (this is in fact a weaker version of our inductive hypothesis). Note that this implies $a_{k+2} \geq 1$. We then have: \begin{align*} a_k&=a_{k+2}-a\cdot a_{k+1}\pm 1\\ &\geq -a \cdot a_{k+1}\\ &> a_{k+1}. \end{align*}So the next term in the sequence is strictly increasing. This completes the induction, which implies that $a_0>a_{n-2}=1$. But we have established that $a_0\in \{-1,1\}$, so this is a contradiction. Combining these cases eliminates the other solutions as desired. $\blacksquare$

We claim that the answers are $(a,b)=(\pm 1, \pm1),(0,\pm1), (\pm2, 1)$. These can be achieved by the polynomials $P(x)=1, x+1, x-1$ respectively. We will prove that these are the only solutions. Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_0$. Notice that $b$ must equal $\pm 1$, since otherwise the constant term $c_0$ cannot be $\pm 1$. This also means that $a_0=\pm 1$. Assume WLOG that $a_0=1$, because otherwise we can negate every coefficient, which doesn’t affect our later argument. Suppose that $b=1$. Then we will prove that if a polynomial exists with $|a|\ge 3$, the coefficients $a_i$ have unbounded absolute value, which would cause a contradiction. We will prove by induction that $|a_{n+1}|>|a_n|$ for all nonnegative integers $n$. The $x$ coefficient of $(x^2+ax+1)P(x)$ is equal to $a+a_1$, and as this must be $\pm 1$, $a_1$ is either $-(a-1)$ or $-(a+1)$, as $|a|\ge 3$, the base case is finished. Notice that the $x^i$ coefficient of $(x^2+ax+1)P(x)$ is equal to $$a_{i-2}\cdot 1+ a_{i-1}\cdot a + a_i \cdot 1.$$This must be equal to $\pm 1$, so $$a_{i-2}+ a_{i-1}a + a_i=\pm 1 \Rightarrow a_{i}= -a_{i-1}a -a_{i-2}+\pm 1.$$This means that \begin{align*} |a_i| &=|-a_{i-1}a -a_{i-2}+\pm 1| \\ & \ge |-a_{i-1}a|-|a_{i-2}|-1 \\ &>a|a_{i-1}|-|a_{i-1}|-|a_{i-1}| \\ &=(a-2)|a_{i-1}|\ge |a_{i-1}|, \end{align*}and the induction is complete. Suppose that $b=-1$. Then we will prove that if a polynomial exists with $|a|\ge 2$, the coefficients $a_i$ have unbounded absolute value, which would cause a contradiction. The case of $|a|\ge 3$ can be solved using the same induction as earlier. If $a=2$, it is easy to prove by induction that all of the terms $a_i$ are positive. We have the relation $$a_{i-2}\cdot 1+2a_{i-1}\cdot - a_i \cdot 1=\pm 1 \Rightarrow a_{i}= 2a_{i-1} +a_{i-2}+\pm 1.$$This means that $a_{i}\ge 2a_{i-1}$ for $i \ge 2$, and this case is resolved. The $a=-2$ case follows from this as well because we can simply replace $x$ with $-x$ to reduce to the $a=2$ case. The proof is complete.

what does it mean when it says $P(x) \in \mathbb{Z}[X]$

This means that $P(x)$ is a polynomial with integer coefficients.

oh ok thank you

The desired ordered pairs $(a,b)$ are those for which $|a| \leq 1$ and $|b| = 1$, plus $a = \pm 2$ and $b = 1$. Constructions are given above, so I won't worry about them. First observe by comparing constant terms that $b = \pm 1$. Call an ordered pair bad if $b = \pm 1$ yet $(a,b)$ is not among the pairs listed. The key observation is the following. Claim. Any quadratic $x^2 + ax + b$, where the pair $(a,b)$ is bad, has a root of magnitude at most $\tfrac 12$. Proof. In the $|a| = 2$ case, compute manually that either $\sqrt 2 - 1$ or $1 - \sqrt 2$ is a root of the quadratic. Hence assume $|a| \geq 3$. The roots of the quadratic are of the form $\tfrac{-a\pm\sqrt{a^2\pm 4}}2$. If $a$ is positive, then the root $t:=-a + \sqrt{a^2\pm 4}$ satisfies \[ \left|-a + \sqrt{a^2 \pm 4}\right| = \frac{4}{a + \sqrt{a^2\pm 4}} \leq \frac{4}{3 + \sqrt 5} < 1, \]so $|t| < \tfrac12$. If $a$ is negative, perform the same analysis with the root $-a - \sqrt{a^2 \pm 4}$. $\square$ From this, we can kill the problem. Suppose the pair $(a,b)$ is bad, and let $t$ be the root of magnitude at most $\tfrac 12$. Then \begin{align*} 0 = |t^n + c_{n-1}t^{n-1} + \cdots + c_1t + c_0| \geq |c_0| - \sum_{j=1}^n|c_jt^j| \geq 1 - \sum_{j=1}^n\frac1{2^j} > 0, \end{align*}contradiction.

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ISL marabot solve Note that $b\in\{-1,1\}$. Also, by the Cauchy bound, the maximum magnitude of a root of the polynomial $x^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0$ is $2$. Case 1: $b=1$. Then one root of the polynomial is $\frac{-a- \sqrt{a^2-4}}{2}$. If $a\ge 3$, then $\left|-\frac{a+\sqrt{a^2-4}}{2}\right|>2$. Another root is $\frac{-a+\sqrt{a^2-4}}{2}$. If $a\le -3$, then $\left|\frac{\sqrt{a^2-4}-a}{2}\right|>2$. So $-2\le a\le 2$. Case 2: $b=-1$. Then one root of the polynomial is $\frac{-a+\sqrt{a^2+4}}{2}$. If $a\le -2$, then $\left|\frac{\sqrt{a^2+4}-a}{2}\right|>2$. Another root of the polynomial is $\frac{-a-\sqrt{a^2+4}}{2}$. If $a\ge 2$, then $\left|-\frac{a+\sqrt{a^2+4}}{2}\right|>2$. So $-1\le a\le 1$. We claim that the solutions are $\boxed{(a,b)\in \{(-2,1), (-1,1), (0,1), (1,1), (2,1), (-1,-1), (0,-1), (1,-1)\}}$. Clearly nothing else can be a solution. It suffices to show that they all work. For the second, fourth, sixth, and eighth, just set $P\equiv 1$. For the first solution, we have $(x^2-2x+1)(x+1)=x^3-x^2-x+1$. For the third, we have $(x^2+1)(x-1)=x^3-x^2+x-1$. For the fifth, we have $(x^2+2x+1)(x-1)=x^3+x^2-x-1$. For the seventh, we have $(x^2-1)(x-1)=x^3-x^2-x+1$.

Suppose such a polynomial exists. Let $r\in\mathbb C$ be a root of $x^2+ax+b$. Then: $$|r|^n=\left|\sum_{k=0}^{n-1}c_kr^k\right|\le\sum_{k=0}^{n-1}|c_k||r|^k=\sum_{k=0}^{n-1}|r|^k=\frac{|r|^n-1}{|r|-1}.$$If $|r|>2$ then this rearranges as $|r|^n(|r|-2)+1\le0$, a contradiction. Thus $|r|\le2$. Since $c_0\in\{-1,1\}$ and $b\mid c_0$, $b\in\{-1,1\}$ also. Then: $$r=\frac{-a\pm\sqrt{a^2\pm4}}2.$$If $b=-1$ and $a\ge2$, then $\frac{-a-\sqrt{a^2-4}}2<-2$, a contradiction. If $b=-1$ and $a\le-2$, then $\frac{-a+\sqrt{a^2-4}}2>2$, a contradiction. If $b=1$ and $a\ge3$, then $\frac{-a-\sqrt{a^2+4}}2<-2$, a contradiction. If $b=1$ and $a\le-3$, then $\frac{-a+\sqrt{a^2+4}}2>2$, a contradiction. It remains to check finitely many cases: If $a=\pm1$ we can use $P(x)=1$. If $a=0$ we can use $P(x)=x-1$. If $a=\pm2$ and $b=1$ we can use $P(x)=x+\frac a2$. These are all of the possible triples.

It is easy to see that $b=1,-1$. If $P(x)$ is constant, we must have $(a,b) = (\pm 1 ,\pm 1)$, and this easily works by letting $P(x)=1$. If $P(x)$ is linear, let it be $P(x)=x+c$. Then the expansion of $(x^2+ax+b)(x+c)$ is $x^3+(a+c)x^2+(ac+b)x+bc$. Since $b = \pm 1$, $ac = -2, 0, 2$. But we also need $a+c= \pm 1$, so we can easily find that $(a,c) = (2,-1), (-2, 1), (0, 1), (0,-1)$. If they are $(2,-1)$ or $(-2,1)$, we need $b=1$, but $0$ works with both values of $b$. We currently have the ordered pairs $(-1, -1), (-1, 1), (1, -1), (1,1), (2,1), (-2, 1), (0, 1), (0,-1)$. Now assume the degree of $P(x)$ is greater than $1$. Let $P(x) = x^{n-2} + d_{n-3} x^{n-3} + \cdots + d_1 x + d_0$. Consider the coefficient of $x^2$ in the product. It is equal to $b d_2 + ad_1 + d_0$. We know $d_0 = \pm 1$ and $bd_2 = \pm 1$. Furthermore, $a+d_{n-3} = \pm 1$, so the maximum value of $a$ is $2$ and least is $-2$. We have already taken all possible cases for $a=-1, 0, 1$ (since $b=\pm 1$ is given), we just need to check whether $(2, -1)$ and $(-2, -1)$ work. We claim that none of the roots of $x^2+ax+b$ can have an absolute value greater than or equal to $2$. Let $x$ be the root. But $|x|^n \geq \max( \sum_{k=0}^{n-1} c_k x^k ) = \sum_{k=0}^{n-1} |x|^k = \frac{|x|^n -1 }{|x|-1}$, and both quadratics $x^2-2x-1$ and $x^2+2x-1$ have a root with an absolute value greater than or equal to 2, so we are done.

We claim that the answer is pairs $(a,b)$ satisfying $|a| \le 1$ and $|b|=1$ or $|a|=2$ and $b=1$. It is obvious that $b=1$ or $b=-1$. These can be verified with $1$, $x+1$, and $x-1$ applied to the right polynomials. Assume $(a,b)$ is not in this set. Note that $|b|=1$. If $|a|=2$ and $b=-1$ we get no solutions. If $a>2$, then $$|\frac{a-\sqrt{a^2 \pm 4}}{2}| = |\frac{2}{a + \sqrt{a^2 \pm 4}}| < \frac{2}{3+\sqrt{5}} < \frac{1}{2},$$so there is a root with magnitude less than $\frac{1}{2}$. A similar argument works for $a<-2$. Taking this root, say $r$, we have that $$0=|\sum_{k=0}^{n}c_kr^k | \ge |c_0| - \sum_{k=1}^{n} |c_kr^k| =\frac{1}{2^n},$$a contradiction so we get the desired polynomials and we are done.

Note that for each root $r$ of $x^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0$, $\tfrac12 < |r| < 2$. Note that $b=\pm 1$. For each $a$, there must exist a root $r$ whose absolute value is \[\frac{|a|-\sqrt{a^2-4}}{2} ~\text{ or } ~\frac{\sqrt{a^2+4}-|a|}{2}\]both of which exceed the limits of $|r|$ when $|a|\ge 3$. Now, the following work: \[(a,b) = (0,\pm 1) : (x^2\pm 1)(x-1) = x^3 - x^2 \pm x \pm 1 \]\[(a,b) = (\pm 1,\pm 1) : (x^2\pm x\pm 1)(x^3+1) \text{ works}\]\[(a,b)=(\pm 2, 1): (x\pm 1)^2(x\mp 1) = (x^2-1)(x\pm 1) \text{ which works}\]However, $(a,b)=(\pm 2, -1)$ has roots with oversized absolute values so no solutions there.

A cute and worthy problem to dedicate my 800th post to! Claim:- for any root $z$ of $x^2+ax+b$ we have $|z| \leqslant 2$ Pf:- FTSOC $|z|>2$ then we have $z^{n}=-\sum_{i=0}^{n-1} c_{i} z^{i}$ we have $|z^{n}|=\left| \sum_{i=0}^{n-1} z^{i} c_{i}\right| \leqslant \left|\frac{z^{n}-1}{z-1}\right|$ hence we have $1 \geqslant \frac{|z^{n}-1|}{|z^{n}|} \geqslant |z-1| \implies 1 \geqslant |z-1|$ also we have $|z|>2$ but also $|z-1| \geqslant |z|-1 > 2-1=1$ which is a contradiction. \[\rightarrow \leftarrow\] So claim follows. $\square$ Now $|b \cdot P(0)|=1 \implies |b|=1 \implies b=\pm 1$ so $z=\frac{-a \pm \sqrt{a^2 \pm 4}}{2}$. Clearly for $|a| \geqslant 3$ we have any one root to be $>2$, hence we have $|a|=0,1,2$ so finally all such $(a,b) \in \mathbb{Z}$ that works are: $a= \pm 1 , b= \pm 1 , P(x) \equiv 1$ $a=0, b =\pm 1 , P(x) \equiv x+1$ $a=-2 , b=1 , P(x) \equiv x-1$ $a=2, b=1 , P(x)=x-1$ $\blacksquare$

Relatively easier problem, I feel like this is accessible even to an algebra student. Note that b is obviously $\pm1$ since otherwise $|c_0|>1.$ On the other hand, we see that for any root r of the polynomial, it must have $1/2<|r|<2$ since otherwise plugging in that root has at least $|\pm{2^n}+...|,$ but $2^n-2^{n-1}-...-1>0$ which would be the minimum value it can take, and analogously for 1/2 where c_0 "outweighs" them. On the other hand, there will be a root that is $$\frac{|a|\pm \sqrt{a^2\mp 4}}{2} \ge 2\iff |a|\ge 3.$$Now it's just a matter of testing, from which we find $$(a,b) = (0,\pm 1) : (x^2\pm 1)(x-1) = x^3 - x^2 \pm x \pm 1,$$$$(a,b) = (\pm 1,\pm 1) : (x^2\pm x\pm 1)(x^3+1),$$and $$(a,b)=(\pm 2, 1): (x\pm 1)^2(x\mp 1) = (x^2-1)(x\pm 1)$$all work, while $(a,b)=(\pm 2, -1)$ has too large of an absolute value of roots in the quadratic. $\blacksquare$

Redacted

Pretty standard by looking at roots. If some root $z$ of the RHS has $|z| \geq 2$, then $$|z|^n > |c_{n-1}z^{n-1}+c_{n-2}z^{n-2}+\cdots+c_0|$$by triangle inequality. So it suffices to find all quadratic polynomials $x^2+ax+b$ which have roots of magnitude $|z|<2$. These work for $(a, b) = (\pm 1, \pm 1) = (\pm 2, 1) = (0, \pm 1)$, which we can all verify to work.

I claim the solutions are $(a,b) = (2,1), (1,1), (0,1), (-1, 1), (-2, 1), (1, -1), (0, -1), (-1, -1)$. We check all of these work: $x^2 + 2x + 1 \cdot (x - 1) = x^3 + x^2 - x - 1$ $x^2 + x + 1 \cdot 1 = x^2 + x + 1$ $x^2 + 1 \cdot x + 1 = x^3 + x^2 + x + 1$ $x^2 - x + 1 \cdot 1 = x^2 -x +1$ $x^2 - 2x + 1 \cdot x + 1 = x^3 - x^2-x + 1$ $x^2 + x - 1 \cdot 1 = x^2 + x - 1$ $x^2 - 1 \cdot x + 1 = x^3 + x^2 - x - 1$ $x^2 - x - 1 \cdot 1 = x^2 -x - 1$ Now to prove nothing else works, we begin with a claim. Claim: All roots, real or imaginary, of the product, must have magnitude less than $2$. Proof: Assume that there exists some $r \ge 2$ such that $r^n = - \sum c_i r^i$. The magnitude of the left side is $|r|^n$, the magnitude of the right side by triangle ineq is at most $\sum |r|^i = \frac{|r|^{n} - 1}{|r| - 1}$, it is easy to check that for $|r| \ge 2$ the numerator of that fraction is less than $|r|^n$ and the denominator is at least $1$, so the fraction is always less than $|r|^n$, so equality can never exist and such $r$ can also never exist. Now clearly we require $b = 1,-1$. In the former case, we can eliminate integer $a > 2$ since $\frac{-a - \sqrt{a^2 - 4}}{2}$ has magnitude at least $2$ for $a > 2$, likewise the identical argument holds for $a < -2$ where we flip the sign on the square root. We can do the same thing for latter case where we analyze the magnitude of $\frac{-a - \sqrt{a^2 + 4}}{2}$, which has magnitude at least $2$ for $a > 1$, and flipping the square root also bars us from having $a < -1$. This leaves only the cases discussed before.

We claim that the only solutions are: $(a, b) = (\pm 1, \pm 1), (0, \pm 1), (\pm 2, 1)$. Note that, $b = \pm 1$. Let $r_1, r_2$ denote the roots of polynomial $g(x) = x^2 + ax+b$. Notice that: $$1 = \left| \sum_{k=0}^{n-1} \frac{c_{k}}{r^{n-k}} \right| \le \sum_{k=1}^{n} \frac{1}{|c|^k} < \frac{1}{|c|-1}$$Thus, $|c| < 2$ and we must have $|a| < 3$. Checking all the remaining cases, we are get these solutions: $(a, b) = (\pm 1, \pm 1), (0, \pm 1), (\pm 2, 1)$ and we are done.

The answers are $\boxed{(\pm 2,1),(\pm 1,1),(0,1),(\pm 1,-1),(0,-1)}$. Clearly it suffices to show it for the absolute value of $a$ because we can just take $x\to -x$ to get the other one. Indeed, \begin{align*} x^2+2x+1\mid&\,\, x^3+x^2-x-1 \\ x^2+x+1\mid&\,\, x^2+x+1 \\ x^2+1\mid&\,\, x^3+x^2+x+1 \\ x^2+x-1\mid&\,\, x^2+x-1 \\ x^2-1\mid&\,\, x^3+x^2-x-1. \end{align*} To prove that these are the only solutions, first note that $|b|\mid |c_0|=1$, so $b\in\{1,-1\}$. Now note that the roots of $x^n+c_{n_1}x^{n_1}+\dots$ all have magnitude less than $2$ (by triangle inequality), so $\left|-a\pm\sqrt{a^2-4b}\right|<4$. The average of these two numbers has to have magnitude less than $4$ then (again by triangle inequality), so $|a|\le 4$. Now we may manually check that the claimed answers are indeed the only ones that satisfy this inequality. $\blacksquare$