I am sure I saw a similar question like this one before but I can't find it now. I tried using "order" but failed. It is quite obvious when $y$ is an even number. The real problem is when $y$ is odd. Is there any easy way to solve this? Thanks.
120797
I am looking for integer solutions of Diophantine equation $m^2+1=5^n$. I found that $m=0,n=0$ and $m=2,n=1$. I could not find any other solutions. I try to prove this but I could not. Could anyone help me to solve this equation?
92881
Find all natural numbers $x,y$ such that   $$3^x=2y^2+1$$  solutions are $(1,1)$, $(2,2)$, $(5,11)$. I found that parity of both is same and If $x$ Is odd it is of the form $4k+1$.
8768
This is problem #6 from Section 1.2 of Ash's Real Variables With Basic Metric Space Topology.   I am asked to show that it is impossible to list the rational numbers in increasing order.   While I know it is possible to list a finite subset of the rational numbers in increasing order, I was wondering if the reason for the impossibility in this case is because there is no least element of $\mathbb{Q}$?   I mean, it's possible to impose an ordering on $\mathbb{Q}$ (correct me if I'm wrong, but it seems possible to compare any two rationals).  But the set $\mathbb{Q}$ is of course, a subset of itself, and the Well-Ordering Principle says that a set $S$ is Well-Ordered only if any subset of $S$ contains a minimal element. Since $\mathbb{Q}$ does not contain a minimal element, it does not appear that it is a Well-Ordered set.   Is this the correct idea as to why it's impossible? If not, could somebody explain to me what the correct idea is and a strategy for proving it?  Thank you ahead of time.
24187
Hello everybody! The problem you see above is a combinatorial problem that I could not solve. :(  Since we are to choose distinct numbers from $\{1,2,3 \ldots K\}$, the numbers must be in increasing order because they are all positive. I think that the difficulty of this problem is in the fact that the Special Sequence can be of any length. We can go about choosing say $n$ elements in $\binom{K}{n}$ ways but again how do we control that? Talking about a recursive solution, I think that is the way to go but I can't get how to do that (that is get a recursive formula for that).  Any help in solving this problem would be much appreciated. Thanks.
26742
It can't be $5$. And it can't be $4.\overline{9}$ because that equals $5$. It looks like there is no solution... but surely there must be?
48302
Show that if $a$ is an odd en integer and $b$ is an even integer then $\gcd(a,b)=\gcd(a,b/2)$.  I understand that since $a$ is not divisible by $2$ but $b$ is, the gcd of $a$ and $b$ also can't be divisible by $2$ but I'm getting stuck using this to help me.
10813
The following is a Homework Question that I've been working on and I would like some feedback on my answer: Prove or find a counterexample: For all real numbers $x$ and $y$ it holds that $x + y$ is irrational if, and only if, both $x$ and $y$ are irrational.  So I've got (or at least like to think I've got) a counter example.  Let $x = 1$ and $y = \sqrt{2}$.   Thus $x$ is rational and $y$ is irrational. Adding $x$ and $y$ gives us $1 + \sqrt{2}$ which cannot be simplified any further (right?) and is an irrational number.  Thereby proving that $x + y$ can be irrational without both x and y being irrational.   Is this okay? Any feedback is greatly appreciated thank you!
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.)
40771
I know that I should use the definition of an odd integer ($2k+1$), but that's about it.  Thanks in advance!
123693
Yes, there is no one who doesn't know this problem.My question is only about curiosity.     $$C(n) = \begin{cases} n/2 &amp;\text{if } n \equiv 0 \pmod{2}\\ 3n+1 &amp; \text{if } n\equiv 1 \pmod{2} .\end{cases}$$   On this problem, I caught something like this.I'm sure, We all realized that.  For example, $n=19$, we have $6$ odd steps.  We know that, even steps are not important, because each even number is converted to an odd number.     $19\Longrightarrow 29 \Longrightarrow 11\Longrightarrow 17 \Longrightarrow13 \Longrightarrow 5 \Longrightarrow 1$   Then, for $n=77$, We have also $6$ odd steps.     $77\Longrightarrow 29 \Longrightarrow 11\Longrightarrow 17 \Longrightarrow13 \Longrightarrow 5 \Longrightarrow 1$   For $n=9$      $9\Longrightarrow 7 \Longrightarrow   11 \Longrightarrow 17 \Longrightarrow 13\Longrightarrow 5 \Longrightarrow 1$   Again we have $k=6$ odd steps.     I want to know / learn / ask, for $k=6$, (Generalized: for any number $k$ ) can we produce a formula(s) to catch all such numbers, which gives the result  $1$?    Thank you!
57696
I am having trouble proving this directly. I know that it is easy to prove by contradiction by assuming both $a$ and $b$ to be odd, but how should I start to try to prove this? This is a homework problem and I am just looking for help getting started, not the full answer.
3005
Some books say that even numbers start from $2$ but if you consider the number line concept, I think zero($0$) should be even because it is in between $-1$ and $+1$ (i.e in between two odd numbers). What is the real answer?
120061
Shouldn't be repeated in any case; if the numbers appeared is exactly the same as another but different orders, that doesn't count as a new set. For example: $11,13,18,19,19,20$ is the same as $13, 18, 19,11,19,20$.
13548
Let $G$ be a finite group having order $n$, prove that $n$ is odd if and only if for each $a\in G$, there is $b\in G$ such that $a=b^2$. I first assume $n$ is odd, then let $a$ be an element in $G$, since $a^n=e$, we have $$a=a^{n+1}=(a^{(n+1)/2})^2$$ which tells $a$ is the square of $a^{(n+1)/2}$.  To prove the converse, I use contraposition, which I assume $n$ is even, then I want to find $a\in G$ such that it is not a square. My intuition is, "assume" we can always find an element $a\in G$ such that the smallest positive integer $k$ satisfy $$a^k=e$$ is $k=n$. Then it is not true that $a$ can be expressed as a square, indeed if $a=b^2$ for some $b\in G$, then the smallest $k$ satisfy $b^k=e$ will be $k=2n$, but this contradicts the smallest $k$ cannot exceed $n$.  My problem is, is the assumption I made always true? If it is true then it should be in the very beginning of the textbook, but I have forgotton it. Any help?  Source: Apostol Analytic Number Theory.
56990
The problem is as stated in the title. I've tried factorization, but it doesn't seem to yield anything. Any help is greatly appreciated. Also please don't use modular arithmetic, thank you.
73735
If $D_{n}$ is the Dihedral group of degree $n$ (order $2n$) where n is an odd number, can I then conclude that the numbers of elements of order 2 in $D_{n}$ is equal $n?$  I suppose there are the following types of elements in $D_{n}$: b, $a^{i}$, $a^{i}b$ for some $0\leq i &lt; n$ .
72355
By contradiction. Suppose that any two vertices have an odd number of common neighbors. For an arbitrary vertex $v$, look at the subgraph induced by the neighbors of $v$. All degrees in this subgraph are odd(why all degrees are odd here?),   and hence the degree of $v$ is even(it is even because of the handshaking theorem?).   Let us count walks of length $2$ beginning at $v$. Their total number is even (because every vertex has an even degree). At the same time, an odd number of these walks come back to $v$, but any other vertex is reached by an odd number of walks, and hence the number of vertices distinct from $v$ must be even—a contradiction.  Please help me visualize this hint from my textbook.