Show that $(x-1)^2$ is a factor of $x^n-nx+n-1$  Here's my approach:- Letting $P(x)=x^n-nx+n-1$ I tried putting $P(1)=0$ but it would only prove for $,(x-1)$  Here's what I can think of, $x^n-1-nx+n=(x^n-1)-(nx-n)=(x-1)(x^{n-1}+x^{n-2}+x^{n-3}................+x+1)-n(x-1)$  Because $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^{n+1}................+xy^{n-2}+y^{n-1})$  $=(x-1)[x^{n-1}+x^{n-2}+x^{n-3}................+x+1-n]$  But I am not able to proceed any further neither I am able to get $(x-1)^2$ as common factor.
121675
Show that $(x-1)^2$ is a factor of $x^n -nx +n-1$   By factor theorem we know that $(x-a)$ is a factor of $f(x)$ if $f(a)=0$.   In this case, $f(x)=x^n -nx +n-1 \implies f(1)=0$  Hence we conclude that $(x-1)$ is a factor. From hereon, how can I say that $(x-1)^2$ is a factor?    Can we approach the problem without calculus approach? This problem was taken from a book of pre-calculus algebra.
97382
If $p(x)$ is a polynomial of degree $n$ and $p(k) = k/(k + 1)$ for $k = 0, 1, 2, \ldots,n$, find $p(n + 1)$.  I'm not looking for a solution, I just want help starting it.    *Edit: Typo in original post,
110473
Let $g_n= {2^2}^n +1 $. Prove $g_0 · g_1 · · · g_{n−1} = g_{n} − 2$.    I'm not sure how to start this proof, if it should be done algebraically or if I should try to use a proof by induction.
84065
I would appreciate a little help in finalizing a proof for the following:  Let $a_n$ be the sequence defined as $a_1 = 1$, $a_2 = 8$, and $a_n = a_{n-1} + 2*a_{n-2}$ when $n \geq 3$. Prove that $a_n = 3*2^{n-1} + 2(-1)^n$.  I decided to use strong induction and show that if the statement is true for $1,...,n$ , then it is true for $n+1$ (going off the fact that the statement is true for $n=3$). I have: $$a_{n+1} = a_n + 2*a_{n-1} = 3*2^{n-1} + 2(-1)^n + 2(3*2^{n-2} + 2(-1)^{n-1}) = 3*2^n + 2(-1)^n + 4(-1)^{n-1}$$  Which is close to the result I want $(3*2^n + 2(-1)^{n+1})$ but the signs are switched for the $(-1)$ terms; therefore I suspect I missed a $-1$ somewhere, but I could not see where the error is.  I appreciate all and any help. Thank you kindly!
98111
Prove that $x^2+y^2=z^n$ has a solution $(x, y, z)$ in $\mathbb{N}$ for all $n\in\mathbb{N}$  I tried to prove this by induction, but couldn't. ( This was probably because the solution for some $n$ isn't necessarily related to the solution for $n+1$)  I can't seem to see any other way other than induction for proving the statement. Any help/hints on solving this problem, or any alternative approach will be highly appreciated :)
46321
The question asks to verify that each equation is true for every positive integer n.   The question is as follows:  $$1+ 3 + 5 + \cdots + (2n - 1) = n^2$$  I have solved the base step which is where $n = 1$.  However now once I proceed to the inductive step, I get a little lost on where to go next:  Assuming that k is true (k = n), solve for k+1:  (2k - 1) + (2(k+1) - 1) (2k - 1) + (2k+2 - 1) (2k - 1) + (2k + 1)   This is where I am stuck. Do I factor these further to obtain a polynomial of some sort? Or am I missing something?
71730
Show that $2$ | $x^{2n} - x^n$, $n \in \mathbb{Z^+}$   Let $x^n = a$ then $\frac{a}{2}$ $\implies a = 2q + r$ such that $r = 0 \vee r = 1$.   So $x^{2n} = (x^n)^2 = a^2 = (2q + r)^2$, which means that $2|[x^{2n} - x^n ]= 2| [(2q + r)^2 - (2q + r)]$. If $r = 1$ we get $(4q^2 + 4q + 1) - 2q - 1 = 4q^2 + 2q = 2(2q^2 +1)$. Thus $2$|$x^{2n} - x^n$, $n \in \mathbb{Z^+}$. Is this reasoning correct? Is it adequate?
86253
Prove that $n&lt;3^n$ where $n \in \mathbb N$,  when $ n=1 $, I have proved it's true. And assumed when $n = p$ , $p&lt;3^p$ is true. Can any body help me in showing that it is true for $n =p+1$
10212
I read the following two sources and I think I understand the structure of proofs involving WOP, but I am stuck on the open-ended part where you prove P(n-1) for the  following problem:    Use the well-ordering principle to show that n^3 - n is divisible by 6.    Any help would be greatly appreciated!   Note: After factoring out (n-1)^3 - (n-1) it took the form of n^3 - 3n^2 + 2n and the only difference between that and n^3 - n is "3n^2 - 3n" which I proved was divisible by 6 via induction. Is there a better way to solve this problem?  Edit: , post is useful however I not supposed to use induction for the main part of the proof.
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.
84323
The question reads     $$U_n = (1+x)^n - 1 - nx$$   Show that $U_2 \geq 0$   Hence or otherwise show that $(1+x)^n \geq 1 + nx$ for all $x \gt -1$.   Obviously the $U_2 \geq 0$ is very easy, I can do that without any trouble but I cannot see how it links to the 2nd part of the question.
42494
Why does:   $$ \sum_{k=0}^{n} k \binom nk p^k (1-p)^{n-k} = np $$ ? Taking the derivative of:   $$ \sum_{k=0}^{n}  \binom nk p^k (1-p)^{n-k} = (1 + [1-P])^n = 1 $$   does not seem useful, since you would get zero. And induction hasn't yet worked for me, since – during the inductive step – I am unable to prove that:   $$ \sum_{k=0}^{n+1} k \binom {n+1}{k} p^k (1-p)^{n+1-k} = (n+1)p  $$   assuming that:   $$ \sum_{k=0}^{n} k \binom {n}{k} p^k (1-p)^{n-k} = np  $$
89936
A polynomial $P$ with integral coefficients satisfies $P(n)&gt;n$ for all positive integers $n$. Every positive integer $m$ is a factor of some number of the form $P(1),\, P(P(1)),\, P(P(P(1))),\dots $. Prove that $P(x)=x+1$.
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.
13550
I don't understand at all why:  \begin{equation} \sum\limits_{n=0}^\infty p_n x^n = \prod\limits_{k=1}^\infty (1-x^k)^{-1} \end{equation}  Where $p_n$ is the number of partitions of $n$. Specifically how can the sum of positive numbers ever be factored into a fraction or factors to the $-1$ power? Is this something purely notational or am I missing something big?  Thanks for all the help!
19878
Let $p, q$ be prime numbers and $n ∈ \mathbb{N}$ such that $p \not\mid n- 1$.  If $p \mid n^q -1$ then show that $q \mid p-1$.   My ideas: I was thinking about the formula $$n^q -1 = (n-1)(n^{q-1} + n^{q-2} +\cdots+1),$$ but now here I do not know how to show that $q \mid p-1$.  Any hints or solution can be appreciated.
105975
Let: $P(x) = a_nx^n + a_{n-1}x^{n-1} + ..... + a_1x + a_0$ where $a_0a_n &lt; 0$ I have to prove that the Polynomial $P(x)$ has at least one positive root how can I prove it? Any ideas?