What is the condition that $$x^2 + x + 1$$ is a factor of $$(x + 1)^n − x^n − 1$$.
2835
I have to find the form of n i.e. whether n is even or odd and whether it is multiple of 2 or 3 such that:  $x^2 + x + 1$ is a factor of $(x+1)^n - x^n - 1$  What I tried:  $$x^2 + x + 1 = (x + 1)^2 - x $$  also,  $$(x+1)^n-x^n-1= \binom{n}{o}x^n + \binom{n}{1}x^{n-1} + ... + \binom{n}{n-1}x + \binom{n}{n}.1 - x^n - 1$$  $$=\binom{n}{1}x^{n-1} + ... + \binom{n}{n-1}x$$  after getting this far I tried with taking x and n common but still couldn't able to find the right direction.  I need extra hints to solve this.
4967
Find all polynomials $P$ such that   $$P(x^2+1)=P(x)^2+1$$
63619
Prove that the polynomial $f_n(x)=nx^{n+1}-(n+1)x^n+1$ is divisible by $(x-1)^2$  where $n\in\Bbb{N_1}$.  My Attempt   I've tried proving it by induction but can't find a way to prove it for $n+1$ after assuming it holds for $n$; I've also checked it graphically, it seems that for all $n$ $f_n$ intersects the axes at $(0,1)$ and $(1,0)$; I've also tried finding a factor $k$ such that $(x-1)^2k=f_n(x)$ for $n=3, n=4$ on WA but with no luck.
52795
Let $n$ be an integer, prove that every prime factor of $12x^2+1$ has the form $6m+1$, where $m$ is an integer.  I've gone as far as knowing that all primes $p&gt;3$ are either $6m+1$ or $6m-1$, how do I use this, not sure whether to even use that.  I know about the Legendre symbol and quadratic reciprocity
114077
How can we show that the assumption $a,b,c&gt;0$ and $a+b+c=1$ implies   $$\sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ca+b} \ge 1+ \sqrt{ab}+\sqrt{bc}+\sqrt{ca}~?$$
38448
Let $x_1, x_2 \in \mathbb R$ and let $(a_0, a_1, \dots, a_n), (b_0, b_1, \dots, b_n)$ be $(n+1)$-tuples of real numbers. Show that there is a unique polynomial of degree at most $2n+1$ such that $$q^{[k]}(x_1)=a_k,$$$$q^{[k]}(x_2)=b_k$$ for $k=0, \dots, n$.   Any hints on how to get started with this exercise?
87297
Let $f(x) = x^2 + ax + b$ be a quadratic polynomial in which $a$ and $b$ are integers. If for a given $n$ , $f(n)f(n+1) = f(m)$ for some integer $m$ then the value of $m$ is ?  Do we have to substitute $x$ by $n$ and $n+1$ and then multiply it? or is there another method to get the solution because the mentioned process will give an extremely large equation.  Answer is $n(n+1)+an+b$
36969
So I am trying to show that $f(x) = x^4-3$ will factor over $\Bbb{F}_p$ when $p\equiv 3\mod 4$, with $p\neq 3$. Specifically, I am trying to show that $f$ will factor into irreducibles such that there is at least one factor of degree 2.  My Work So Far  So if I assume that it can be factored, we would have $$ x^4-3 = (x^2+ax+b)(x^2-ax+d) = x^4 + (b+d-a^2)x^2 + a(d-b)x + bd.$$ Because we want the linear term to be 0, we need either $a=0$ or $d-b = 0$. Now if $a=0$ we get that $b=-d$, so $-b^2 = -3$. That is, $3$ is a square. If $d-b = 0$, then we would have $b^2 = -3$, so $-3$ is a square. Because $-1$ is not a square when $p\equiv 3\mod{4}$, we know that one of $-3$ or $3$ must be a square, but not both, so $a=0$ or $d=b$, but not both.  If $a=0$, that is $3$ is a square, then we have $f(x) = (x^2 +b)(x^2-b)$. Then one of $b$ or $-b$ is a square, but not the other, so $f$ will factor into two linear terms and a quadratic. So this case works.  The trouble I am having is with the second case. If $d=b$, then we have  $$ f(x) = (x^2 + ax + b)(x^2 -ax +b)$$ with $2b = a^2$. This is where I get stuck. How do I know that such an $a$ can exist?  Edit: If there is a better approach to solving this, that would be great to know as well. I haven't been taught any of the relevant number theory to solve this, so I'm kind of out of my element here.
106971
I know that you can't factor a real polynomial into $\Pi_{i=1}^N(x-a_i)$ in general. But is it possible to factor every real finite polynomial into this form:    $(\Pi_{i=1}^N a_ix^2 + b_ix + c_i) (\Pi_{i=1}^Mx-d_i)$  where the second term is possibly empty?  This is at least true for polynomials with odd powers because they must have at least one real root. (My reasoning was false, edit see below) So I wonder if it is also true for even powers?    Edit: Berci has pointed out a lapse in my reasoning. For an odd power, once you factor out the first root, it is conceivable that you may hit an even-degree polynomial that is irreducible.
49660
I wanted to prove the following question in an elementary way not using Bertrand postulate or analytic estimates like $x/\log x$. The question is  $$ p_{n+1}^2&lt;p_1p_2\cdots p_n,\qquad(n\geq4) $$  I made the following argument. Does anyone have some opinion or simpler ideas to complete.  We consider two cases: Case 1: $N=p_1p_2\cdots p_n-1$ is composite: then there will be a prime factor  $q\leq\sqrt{N}$, and of course $q$ is not any of the $p_i$'s. therefore $p_{n+1}\leq q\leq\sqrt{N}$. So $p_{n+1}^2&lt;N+1$.  Case 2: $N=p_1p_2\cdots p_n-1$ is prime. Then???
81312
Let $n &gt; 0$ and let $a_{n-1}, ..., a_0$ be integers. Prove that, if $x^n + a_{n-1}x^{n-1} + ... + a_0 = 0$ has a rational root, then such root is an integer.   I understand that posting questions without showing your work is frowned upon, but I don't know where to start with this one.. thanks in advance.
116889
Prove $x^{rs}-1=(x^s-1)(x^{s(r-1)}+x^{s(r-2)}+...+x^s+1)$  I can see that this is true for simple results such as $x^6$ and $x^9$  How can I prove that this is true in general.
23125
I aim to show that $$(n+1)^n&lt;n^{(n+1)}$$ for all $n \geq 3$. I tried induction, but it didn't work. What should I do?
60101
Let $n \in \mathbb N^+$. Show that if $x_1, x_2, ... , x_n$ are $n$ real numbers such that $-1 \le x_i \le 0$ for each $1 \le i \le n$, then $$(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + x_1 + x_2 + ... + x_n$$  I am using a proof by induction, but I am unable to complete it. I would appreciate any tips on how I could finish the proof.   Incomplete Proof Let $P(n)$ be the proposition that $(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + \sum_{i = 1} ^ n x_i$. When $n = 1$, LHS = RHS = $1 + x_i$. Clearly $1 + x_i \ge 1 + x_i$ and so $P(1)$ is true. Assume that $P(k)$ is true, i.e. $(1 + x_1)(1 + x_2)...(1 + x_k) \ge 1 + \sum_{i=1}^k x_i$. Case 1 Suppose $x_{k + 1}  = -1$. Then $1 + x_{k + 1} = 0$, hence LHS = $(1 + x_1)(1 + x_2)...(1 + x_k)(1 + x_{k+1}) = 0$. RHS = $1 + x_1 + x_2 + ... + x_k - 1 = x_1 + x_2 + ... + x_k$ which is always nonpositive since each $x_i$ is nonpositive. Hence RHS $\le 0$ and so LHS $\ge$ RHS. Case 2 Suppose $x_{k+1} = 0$. Then $1 + x_{k+1} = 1$, hence multiplying it to LHS does not change anything. Also, adding $x_{k+1}$ to RHS does not change anything. Therefore, LHS $\ge$ RHS still holds. Case 3 Suppose $-1 &lt; x_{k+1} &lt; 0$. Then $0 &lt; 1 + x_{k+1} &lt; 1$. Adding $x_{k+1}$ to RHS makes the RHS smaller since $x_{k+1}$ is negative. If LHS $= 0$, then multiplying $1 + x_{k+1}$ to LHS does not change anything, and so LHS $\ge$ RHS holds. If LHS > $0$, then multiplying $1 + x_{k+1}$ to LHS makes the LHS smaller since $0 &lt; 1 + x_{k+1} &lt; 1$. In this case LHS $\ge$ RHS does not necessarily hold? ...
81826
$(x+2)(x-1$)  Totally forgot how to do it. Isn't this the rule:  $(a+b)(c+d)$  $a*c +a*d +b*c +b*d$  The answer I'm getting is $x^2-x+2x-2$ which is wrong. Please help me out, thanks!
29954
This problem was in IMC 2010.  Assuming $x,y,z\in [-1,1]$, suppose that $$1+2xyz\geqslant x^2 + y^2 + z^2$$ Can we infer from this that $$1+2(xyz)^n\geqslant x^{2n} + y^{2n} + z^{2n}$$ for any positive integer $n$?
11249
I have an equation containing factorials  here  $$(k+1)!+(k+1)(k+1)!$$  yet I am having a hard time understanding how to simplify it using algebra. A simple search on wolfram gets me a reduced form of   $$(k + 2)! $$  This would be a great refresher to such problems, sadly I don't know the elementary operations to reduce it.