Question: Suppose $P(x)$ is a polynomial with $P(2)=2017$ and $P(5)=2002$. If it is given that $P(x)=0$ has exactly one integer root, find that root.  My approach: I tried solving the problem by considering that $P(x)\in\mathbb{Z}[x]$ and that $$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0.$$ Then $P(5)\equiv a_0 \pmod 5\implies 2002\equiv a_0 \pmod 5\implies a_0\equiv 2\pmod 5.$ Again, $P(2)\equiv a_0 \pmod 2 \implies 2017\equiv a_0 \pmod 2\implies a_0\equiv 1 \pmod 2.$ Thus, we have $$\begin{cases}a_0\equiv 1\pmod 2,\\a_0\equiv 2\pmod 5.\end{cases}$$ By solving the same we can conclude that $a_0\equiv 7 \pmod {10}.$ Also if $a$ is the only integer root of $P(x)$, then there exists $Q(x)\in \mathbb{Z}[x]$ with $\deg (Q(x))=n-1$ such that $$P(x)=(x-a)Q(x).$$ Thus, we have $$(2-a)Q(2)=2017,\text{ and }(5-a)Q(5)=2002=2\times 7\times 11\times 13.$$ After this I couldn't make any significant approach. I also understand that I shouldn't have made the assumption of $P(x)\in\mathbb{Z}[x]$, but that's the best I could think of. Can anyone help me progress (initially a small hint would be nice)?
122716
$P(x) = 0$ is a polynomial equation having at least one integer root, where $P(x)$ is a polynomial of degree five and having integer coefficients.  If $P(2) = 3$ and $P(10)= 11$, then prove that the equation $P(x) = 0$ has exactly one integer root.  I tried by assuming a fifth degree polynomial but got stuck after that.  The question was asked by my friend.
94686
Suppose that $P(x)$ is a polynomial of degree $n$ such that $$P(k)=\dfrac{k}{k+1}$$ for $k=0,1,\ldots,n$. Find the value of $P(n+1)$.    I could not relate this question with this one , so maybe it is not a duplicate  I have absolutely no idea, how the function became a polynomial and how to approach it. Please help!
121582
Fix a nonnegative integer $n$. Let $P(x)$ be a polynomial of degree $n$ (over the real numbers) such that for all $k\in\left\{0,1,\ldots,n\right\}$, we have $P(k)=\dfrac{k}{k+1}$. Find $P(n+1)$.   My attempt:  I consider a new polynomial $Q(x)=(x+1)P(x)-x$. This $Q$ is a polynomial of degree $\leq n+1$ satisfying: $Q(k)=0\quad\forall k\in\{0,1,2,\ldots,n\}$. Therefore $Q(x)=\lambda x(x-1)\cdots(x-n)$ for a certain real $\lambda$. Hence, $Q(-1)=\lambda \cdot (-1)^{n+1}(n+1)!$, but also $Q(-1)=(-1+1)P(-1)+1=1$. Comparing these, we get $\lambda \cdot (-1)^{n+1} (n+1)! = 1$, thus $\lambda=\dfrac{(-1)^{n+1}}{(n+1)!}$. Since $Q(n+1)=\lambda(n+1)!$ then $Q(n+1)=(-1)^{n+1} $, and from $P(x) = \dfrac{Q(x)+x}{x+1}$ we deduce: $$P(n+1)=\dfrac{(-1)^{n+1}+n+1}{n+2}.$$ I doubted that the answer were $P(n+1)=\frac{n+1}{n+2}$ but the term $(-1)^{n+1}$ wasn’t expected. Is there any mistake here? Does someone have an alternative proof?
53742
Let $A \in \mathcal{M}_{n}(\mathbb{Q})$ such that $A^{2017}=I_n$. Is it always true that $$A=I_n?$$   When $n \leq 2015$, I proved that this result must hold using the minimal polynomial and the fact that $2017$ is prime. I suspect that when $n \geq 2016$, this equality is not always true, but I can't find any matrix other than $I_n$ with that property, which also has only rational entries.  I also thought of an approach with eigenvalues, but since they are all complex roots of $2017$, I am not sure how to make use of them here.
100835
Let $p(x)$ be a polynomial with integral coefficients. Let $a$, $b$, $c$ be three distinct integers such that $p(a) = p(b) = p(c) = -1$. Find the number of integral roots of $p(x)$.
115324
Consider the polynomial   $$ P(x) = x^{n} + a_{n-1}x^{n-1} + ... +a_{1} x + a_{0} $$  with only real roots and coefficients. If $R$ is the smallest bound such that $|x_{i}| \le R$ (it is the bound for the largest root), $P(x_{i})=0, i=0,1,...n-1$, then for every conjugate $\frac{1}{p} + \frac{1}{q} = 1, p,q &gt;1$, prove that  $$ R &lt; \left( 1 + A_{p}^{q} \right)^{1/q}, A_{p} = \left( \sum |a_{i}|^{p} \right)^{1/p}$$    Attempt  We have   $$ P(x_{0}) = P(x_{1}) = ... = P_(x_{n-1}) = 0 $$  sum all of them to get  $$ nx^{n} + a_{n-1} \sum x_{i}^{n-1} + ... + a_{1} \sum x_{i} + n a_{0} = 0$$    Let us start from easy case, $n=2$:  with only two roots $x_{0},x_{1}$ then we have $-x_{0} - x_{1} = a_{1}$, and $ a_{0} = x_{1} x_{0}$. We wish to prove   $$ R &lt;  \left[1 + \left( |x_{1}x_{0}|^{p} + |x_{1} + x_{0}|^{p}   \right)^{q/p} \right]^{1/q} $$  ... Clearly we must have $R = max(|x_{1}|, |x_{0}|)$, WLG letit be $|x_{0}|$. Let us work backwards:  $$ |x_{1}+a_{1}| + |\frac{a_{0}}{x_{1}}| &lt; 2\left( 1 \right)^{1/p} \left[1 + \left( |x_{1}x_{0}|^{p} + |x_{1} + x_{0}|^{p}   \right)^{q/p} \right]^{1/q} $$  $$ |x_{1}+a_{1}| + |\frac{a_{0}}{x_{1}}| &lt; \left( (2^{q})^{p} + (2^{q})^{p} \right)^{1/p} \left[1 + \left( |x_{1}x_{0}|^{p} + |x_{1} + x_{0}|^{p}   \right)^{q/p} \right]^{1/q} $$  How to continue..?
88800
Let $P\in\mathbb{R}[X]$ be a positive polynomial.  I want to show that there exists $A,B\in\mathbb{R}[X]$ so that $P=A^2+B^2$    $\displaystyle P=a\prod_{k=0}^q(x-a_k)\prod_{k=0}^{2p}(x-b_k)=a\prod_{k=0}^q(x-a_k)\prod_{k=0}^p(x^2-2\Re(b_k)+|b_k|^2)$ where $a$ is the leading coefficient, $a_k$ are the real rooks, $b_k$ are the complex roots.  What can i do now ? I know that the identity $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$ should come in handy, but I can't figure out :  1) How to show that the product real roots is a square  2) How to use the above identity for the complex roots.  If I can solve those two points, then I can solve the problem.
19236
Show that the polynomial $X^5 + X^3 + \bar{1}$ in $(\mathbb{Z}/2\mathbb{Z})[X]$ is irreducible. (Hint: if it were reducible, it would either have a root or be of the form $g(X) \cdot h(x)$, where deg$g(X) = 2$ and deg$h(X) = 3$) Recall that $\bar{a}$ is shorthand for the coset $a + 2\mathbb{Z} \space $in$ \space (\mathbb{Z}/2\mathbb{Z})[X]$ (for any $a \in \mathbb{Z}$)  This question is worth $7$ marks out of a possible $75$, this is my solution..  $f(X) = X^5 + X^3 + \bar{1}$ is irreducible in $\mathbb{Z}/2\mathbb{Z}[X]$ iff it has no roots in $\mathbb{Z}/2\mathbb{Z}$. Since $f(\bar{0}) = \bar{1}$ and $f(\bar{1}) = \bar{1}$, it follows that $f(x)$ is irreducible in $\mathbb{Z}/2\mathbb{Z}[X]$.  Would this be a sufficient answer? am I missing something?
61377
I need to solve this equation:  $x^2 = 1$ in the ring $\mathbb{Z}/n \mathbb{Z}$.  I know  that  $(n - 1)(n - 1) \equiv 1 \pmod n$, in general $(n-a)^2 \equiv a^2 \pmod n$.  I also know that for $p$ prime all elements of $\mathbb{Z}_p$ are invertible and here $1$ and $p-1$ are their own inverses.  Could you tell me what other solutions there are to this equation?
61020
Can some of you help me with my homework?     I had to count the irreducible, quadratic, monic polynomials in $\mathbb{F}_p[X]$ for arbitrary $p$.    I will show you what I tried myself.  Research effort  First of all the set $\{X^ 2 + bX +c : b, c \in \mathbb{F}_p \}$ contains $p^2$ elemenst. We need to substract the amount of elements that is reducible. These elements are exactly the set $A = \{(X+c)(X+d) : c, d \in \mathbb{F}_p \}$. There are $p(p-1)$ pairs of $(a,b) \in \mathbb{F}_p \times \mathbb{F}_p$ such that $a \neq b$, and there are $p$ pairs such that $a=b$.     When $a$ and $b$ are unequal, whe know that $a$ and $b$ can be swapped without changing anything to the polynomial $(X+a)(X+b)$, so I obtained that $|A| \leq p +(p-1)- \frac{1}{2}(p-1)=p-\frac{1}{2}(p-1)$.  In other to find out these are the elements, or that there are still less, I thought I had to assume that $(X+c)(X+d) = (X+\gamma)(X+\delta)$, to deduce some information about the realation between $(c,d)$ and $(\gamma, \delta)$, but this led to taking square roots in $\mathbb{F}_p$, and I am not familiar with that.  Could you please provide me some more information?
107169
Question: Let $p$ and $q$ be distinct odd prime numbers.  By considering primitive roots, we need to show $\exists c\in\mathbb{Z}$ with the property that:  $\bullet$ $c^n\equiv 1\pmod{p}$ whenever $n$ is a multiple of $p-1$, and   $\bullet$ whenever $n$ is not divisible by $q-1$, then $c^n\not\equiv 1\pmod{q}$  I am only interested with the beginning of this answer.  Answer: We let $c$ be a primitive root modulo $q$.  We are going to use Fermat's little theorem so we need $(c,p)=1$  If $(c,p)\neq 1$, then we have $p\mid c$, but then we have $(c+q, p)=1$  and since $c+q$ is a primitive root modulo $q$, we can replace $c$ with $c+q$ for $(c,p)=1$ for using Fermat's little theorem.  I understand the general method and argument, but...  Contention: Why is $c+q$ still a primitive root modulo $q$?  When is this generally true? Is there an identity for when things remain a primitive root?  Why is $(c+q)^{\phi(q)}\equiv 1\pmod{q}$?
11804
The problem statement is:  Part(a)  Assume P(z) is a non-constant polynomial with all of its roots in some half plane H.  Show that the derivative P'(z) must also have all of its roots in H.  Part (b)  Show that the roots of P' always lie within the closed convex hull of the set of roots of P.  The hint given is:   After obvious reductions of the problem, look at the behavior of $\frac{P'}{P}$ on the complementary half-plane.  My thoughts:   So I think that, whatever half-plane H that we have to work with, we can always just look at $P(e^{i\theta}z)$, which has its zeros in the UHP, for a suitable rotation by angle theta.    Now we can just show that $P'(e^{i\theta}z)e^{i\theta}$ also has its roots in the UHP -- and then undoing the rotations shows that both P and P' had its roots in the same half-plane H.  Using the hint regarding looking at the complementary half-plane, we then want to show that P' does not vanish in the LHP (we already know P doesn't vanish in the LHP, by assumption.)    Again following the hint we look at P'/P in the LHP and show that it does not vanish.  I am stuck at this point.  How can I proceed?  And regarding part(b) what is a "convex hull"?  I am guessing that the answer to part (b) is immediate, after figuring out part(a).  Any ideas are welcome.  Thanks,
124206
Let $r$ be a root of the polynomial $p(x)=(\sqrt{3}-\sqrt{2})x^3 + \sqrt{2}x-\sqrt{3}+1$. Find another polynomial $q(x)$, with all integer coefficients, such that $q(r)=0$.
90790
There is a question I have based on the fact: If you take a quadratic polynomial with integer coefficients, take the set $\{1,2,3,4,5,6,7,8\}$, make a partition $A=\{1,4,6,7\}$, $B=\{2,3,5,8\}$, and then evaluate the polynomial with the elements of each set you got an equality.  If now with a cubic polynomial and the set $\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16\}$, how to pick a suitable partition that accomplish the same fact? I mean is there a general formula for bigger degree?
18049
Suppose $P(x)$ is a polynomial of degree $2012$ and $P(x) = 1/x$ when $x$ takes the integer values $1\cdots2013$ (inclusive). What is the value of $P(2014)$?   I get $1/1007$ but I'm not sure if it's right, and my method is very inelegant.
2916
Consider the root of the polynomial $p(x) = x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots+a_1x -1$.    Suppose that $p(x)$ has no roots in the open unit disc in a complex plane and $p(-1)=0$. Show that $p(1)= 0$.
51577
I don't think I fully understand.  Let's say there is a root $x_0 \in K=\mathbb{F}_p(t)$, where $p$ is a prime number.  Then $x_0 = \frac{P(t)}{Q(t)}$ for some polynomials $P,Q \in \mathbb{F}_p[t]$. We can assume $\gcd(P,Q)=1$  and $x_0^p-t= \frac{(P(t))^p}{(Q(t))^p}-t= \frac{P(t)^p-tQ(t)^p}{Q(t)^p} =0$, so coefficients of $P(t)^p, tQ(t)^p$ must be identical, which contradicts $\gcd(P,Q)=1$,   hence such $x_0$ does not exist and the polynomial has no root in $K$.   am I about right?    anyway, I would appreciate an explanation about $\mathbb{F}_p(t)$, what is $t$? what is the meaning of a variable which does not belong to any specific "world"? I cannot use $t$ as if it was a member of $\mathbb{F}_p$ and hence cannot assume $t^{p-1} = 1 \pmod p$...