prove $(\sqrt x+\sqrt {x+1})^n$ where n is an odd natural number can always be written as $\sqrt y + \sqrt{y+1}$ while y is also a natural number . I have tried using induction by proving the question when n is an even integer  which is easy,then  I tried reducing $(\sqrt x+\sqrt {x+1})^n$ to  $(\sqrt z+\sqrt {z+1})$.$(\sqrt x+\sqrt {x+1})$ in my induction process  but I hit a hard wall here and don't seem to figure out a  way to write these two parentheses in form of a unified $\sqrt y + \sqrt{y+1}$ . Should I change my approach? Is there any other way to prove this ?
121073
If $m,n\in N$ Prove that there is such a positive integer k, such that $(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}$ I attempted to solve this question using binomial coefficients, saying: $(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}=\sum\limits_{a=0}^n {n\choose a}*\sqrt{m}^{n-a}*\sqrt{m+1}^a$ and from here I was thinking that I had to do something with $\sqrt{m}^{n-a}$ with $\sqrt{k}$ and do something else with $\sqrt{k+1}$ and $\sqrt{m+1}^a$. Unfortunately I couldn't think of what to do with this circumstance. I then thought of maybe taking $(\sqrt{k}+\sqrt{k+1})^2=(\sum\limits_{a=0}^n {n\choose a}*\sqrt{m}^{n-a}*\sqrt{m+1}^a)^2$, but this immediately over-complicated the question. Could you please explain to me how to solve this question and how to solve similar questions in the future?
67186
I have to do proof of $(a+b)^n$ and $(a+b)^{n+1}$ with mathematical induction.  I finished the first one $(a+b)^n = a^n + na^{n-1}b+\dots+b^n$.  I however have trouble with the second one, I don't know what to start exactly. Any help/hints?
6512
Prove that if $n&gt;1$ is odd, then $1^n+2^n+3^n+\ldots+(n-1)^n$ is divisible by $n$.   I'm pretty sure I need to prove this using induction but I am unsure as to how to go about it.
115709
Proving by induction. We'd like to show that $2 + 4 + 6 + \cdots+ 2n = n(n + 1)$.  A nice way to do this is by induction. Let $S(n)$ be the statement above. An inductive proof would have the following steps:  Show that $S(1)$ is true.  Show that if $S(1),\ldots,S(k)$ are true, then so is $S(k + 1)$.    This question is really starting to bug me, am I doing something wrong or is the equation wrong for that series? It only seems to work for $S(1)$ but after that it does not give the correct series. The $n(n+1)$ series is $2, 6, 12,\ldots.$ I'm really confused can someone please nudge me in the right direction (I know how proofs by induction works) I'm just having problems with this one in particular.
118344
Show that $\sqrt{1+n(n+1)(n+2)(n+3)}$ is a whole number for all whole numbers $n$. I can see that there are four consecutive numbers, meaning that the expression can be written as $\sqrt{1+24m}$ Also, it is easy to see that the expression is true for $n=1$ but I can't get the induction to work...
86024
There is , but the definition of "even" and "odd" that I am using uses integers instead of just natural numbers; i.e.,   An integer $n$ is even iff there is some integer $k$ such that $n=2k$. An integer $n$ is odd iff there is some integer $k$ such that $n=2k+1$.   Here is what I have so far:  First we show that an integer $n$ is even or odd. We first use induction on the positive integers. For the base case, $1=2\cdot0+1$ so we are done. Now suppose inductively that $n$ is even or odd. If $n$ is even, then $n=2k$ for some $k$ so that $n+1=2k+1$ (odd). If $n$ is odd, then $n=2k+1$ for some $k$ so that $n+1=2(k+1)$ (even). This closes the induction, so every $n\in\mathbf{Z}^+$ is even or odd.  Now we show every $n\in\mathbf{Z}^-$ is even or odd. Let $n\in\mathbf{Z}^-$. Then $n=-k$ for some $k\in\mathbf{Z}$ (I think this follows immediately from most definitions of the integers.). Suppose $k$ is even. Then $k=2j$ for some $j$ so that $n=-k=-2j=2(-j)$ (even). Now suppose $k$ is odd. Then $k=2j+1$ for some $j$ so that $n=-k=-(2j+1)=-2j-1=-2j-1+1-1=-2j-2+1=2(-j-1)+1$ (odd).  For $0$, note that $0=2\cdot0$ (even).  Now we show that $n\in\mathbf{Z}$ cannot be both even and odd. Suppose for the sake of contradiction that $n\in\mathbf{Z}$ is both even and odd. Then there are integers $k,j$ such that $n=2k=2j+1$. This implies that $2(k-j)=1$ (like in the referenced question). So we must show that $1$ cannot be even in order to complete the proof.  This is where I am having trouble. I know that if I let $f:\mathbf{Z}\to\mathbf{Z};x\mapsto2x$ be a function, then $f$ is increasing so since $f(0)=0$ and $f(1)=2$ and $0&lt;1&lt;2$, there is no integer $m$ such that $f(m)=1$. But this seems complicated so I was wondering if there was an easier way to do this.  So my real question is: how can I show that $1$ is not even?  (This is not homework.)
11556
$n$ is a natural number. I’ve tried to turn this problem to this: Prove that: $n(n+1)(n+2)(n+3)+1 $is always a perfect square. So: $$n(n+1)(n+2)(n+3)+1 =x^{2}$$ And i’ve expanded the whole equation but it didn’t help me a lot.
107
Prove that for all natural number $x$ and $n$, $x^n - 1$ is divisible by $x-1$.  So here's my thoughts: it is true for $n=1$, then I want to prove that it is also true for $n-1$  then I use long division, I get:  $x^n -1 = x(x^{n-1} -1 ) + (x-1)$  so the left side is divisible by $x-1$ by hypothesis, what about the right side?
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?
17436
Here, we suppose that $x,y\in\mathbb{R}$ and that $x^n=y^n$, where $n$ is odd. I want to prove that $x=y$.  Maybe we can use that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$  So, it suffices to show that if $x^n=y^n$ then $x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1}\neq 0$  Any hint?
44709
I'm trying to prove using induction that $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ whenever $n$ is a non-negative integer. So far, I have:   Base case: $P(1) = (1) + (8) + (27) = 36, 36$ can be divided by $9$ so the base case is valid Inductive step: let $P(n)$ be the statement $9$ divides $n^3 + (n+1)^3 + (n+2)^3$. Assume $P(k)$ is true, so $9$ divides $k^3 + (k+1)^3 + (k+2)^3$.   And this is where I'm stuck. I'm not sure how to demonstrate that $9$ divides $P(n)$ when $n = k+1$. If someone could step me in the right direction that would be awesome. Thanks!
66163
I ran across this problem in some old notes, and I frustratingly can't figure out how to do it   Let $a_i$ and $b_i$ be sequences of natural numbers, use induction to show  $\sum_{i=1}^n (a_ib_i)^{1/2} \le (\sum_{i=1}^n a_i)^{1/2}(\sum_{i=1}^n b_i)^{1/2} $  Obviously this is trivial to show for n=1. I can't make much progress on n+1. I've tried various tactics, squaring both sides etc.   Any hint or help would be appreciated. Thanks.
73759
Prove that $\sqrt{n} &gt; \ln n$ for all $n \in \mathbb{N}$.  I need to use this fact for one of the proofs that I am working on. However, I am having trouble proving this. I tried induction but don't really know how to do it. Can someone help me with this?
91047
I'm having some trouble proving the following statement using mathematical induction:   $$\frac{1}{2}n^{\frac{3}{2}} \leq \sqrt{1} + \sqrt{2} + \sqrt{3} + \sqrt{4} + ... + \sqrt{n}      ,\text{ (for all sufficiently large n)}$$   I'm sort of confused because $n^{\frac{3}{2}}$ = $n  n^{\frac{1}{2}}$ which means that there are n of the largest possible root values on the left hand side while the right hand side has values like $\sqrt{1}$, $\sqrt{2}$, etc. As n grows sufficiently large wouldn't the left hand side outgrow the right hand side no matter what the constant multiple is? If I'm thinking about this correctly just let me know, if not, any help would be appreciated.
117827
Establish that if $a$ is an odd integer, then for any $n\ge1$ $$a^{2^n}\equiv1\pmod{2^{n+2}}.$$   [Hint: Proceed by induction on $n.$]   This is a problem from Elementary Number Theory by David M. Burton.  For $\underline{n=1},$ this statement is essentially saying that $$a^2\equiv1\pmod{2^3=8},$$ which is true because $a$ is odd. Suppose that this statement is true for $\underline{n=k}$. Then,$$a^{2^k}\equiv 1\pmod{2^{k+2}}.$$ Squaring this congruence, I arrived at $$a^{2^{k+1}}\equiv1\pmod{2^{k+2}},$$ but how do I change the modulus, i.e., $\pmod{2^{k+2}}\to\pmod{2^{k+3}}?$
2067
It is well known that $\sqrt{2}$ is irrational, and by modifying the proof (replacing 'even' with 'divisible by $3$'), one can prove that $\sqrt{3}$ is irrational, as well.  On the other hand, clearly $\sqrt{n^2} = n$ for any positive integer $n$.  It seems that any positive integer has a square root that is either an integer or irrational number.        How do we prove that if $a \in  \mathbb N$, then $\sqrt a$ is an integer or an irrational number?      I also notice that I can modify the proof that $\sqrt{2}$ is irrational to prove that $\sqrt[3]{2}, \sqrt[4]{2}, \cdots$ are all irrational.  This suggests we can extend the previous result to other radicals.        Can we extend 1? That is, can we show that for any $a, b \in \mathbb{N}$, $a^{1/b}$ is either an integer or irrational?
37910
$m,n$ are integers bigger than 0.   I need to prove:  $$\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}} \geq 1$$   I tried to multiply both sides by the common factor and raise both sides to the power of $m*n$ but it did not work for me, and I have no other idea how to proceed.  I don't want the full solution, but just a hint.  Thank you.