I want to prove that $(2,4)$ is the only natural solution to $x^y=y^x$. I tried to prove by defining $d=\gcd(x,y), x=d*s,  y=d*r$ for some $s$, $r$. Using some algebra, I got that s must be equal to 1. I don't know how to continue from here.    My steps: ${ds}^{dr}={dr}^{ds} -&gt;$ $d^rs^r=d^sr^s -&gt;$ $d^{r-s}s^r=r^s -&gt;$ there for $s|r^s$, and because $\gcd(s,r^s)=1$ must be $s=1$.
18865
We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($x\neq y$) which satisfies the equality $x^y = y^x$?
30991
If $x^{2}-x\in Z(R)$ for all $x\in R$, then $R$ is commutative.  I need to proof this theorem and I have something like this below. However, I do not know how to continue this proof.   $(x+y)^{2}-(x+y)=x^{2}+xy+yx+y^{2}-x-y=(x^{2}-x)+(y^{2}-y)+xy+yx\in Z(R)$  so $xy+yx\in Z(R)$
7518
This is a problem set I have, it's not a homework but it's very important practice...  Send me some hints please, I don't want an answer I need to get it by myself but I'm failing miserably...   The problem is:     Prove that for all natural numbers $n$,  $4^n &gt; n^3$
7574
Find all integer solutions to $x^2+4=y^3$.  Some obvious solutions are $(x,y)=(\pm2,2)$. Are these the only ones?
55457
I need to find the gcd of two polynomials: $f(x) = x^4+1$ and $g(x)=x^2-1$ using the Euclidean algorithm.  Wolfram shows that the gcd is equal to $1$, but for some reason I don't get the same answer.   First I divided $f(x)$ by $g(x)$ and got that the remainder is $2$. Then, I divided $g(x)$ by the remainder, $2$, and got a remainder of zero, hence concluding that the gcd is $2$.   What am I doing wrong?
82158
Proof  of identity $\sqrt {xy} = \sqrt x \sqrt y$ for $x,y \in \mathbb R^+$  I've been looking at the stated identity, which makes sense in $\mathbb R^+$ but fails in $\mathbb R$, since $\sqrt {-1 \cdot -1} \neq \sqrt {-1} \sqrt {-1}$.  How does one prove this identity ?  Suppose we have $x,y \in \mathbb R^+$ then $\sqrt{xy}^2 = xy = (\sqrt x \sqrt y)^2$.  Idea 1: The square-root function is bijective (monotonic increasing) and has inverse $X^2$. This in turn implies $\sqrt {xy} = \sqrt x \sqrt y$  Idea 2: The equation $X^2 = xy$ has at most $2$ solutions in $\mathbb C$, which implies $\sqrt x \sqrt y$ must be equal to $\pm \sqrt {xy}$.  Are these ideas rigorous enough ? Is there some simpler way of proving this ?
45386
The proof I have starts of with $\;xy=0\;$ in a field.   Then $x^{-1}$ exists because it is a field.  Then $x^{-1} xy=x^{-1} 0$.  Therefore $y=0$.  But surely if an integral domain can not have any zero divisors, how can we end the proof by saying $y=0$? Surely then y is a zero divisor and hence the field is not an integral domain?
119571
I am investigating solutions to Fermat's equation $$x^n + y^n = z^n$$ with $x,y,z$ in the Gaussian integers, excluding integers and pure imaginaries.  I have found out that there are only trivial solutions for the $n=3$ and $n=4$ cases, e.g. .   I would be grateful if you let me know of the current status or if it is already a theorem.
29309
If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$   This seems clear, but I don't know how to prove this..  I was trying to show this by induction such that if $a^{n+1}$ = $rs$ and $b^{n+1}$ = $rt$, then $s,t$ are divisible by $a,b$ respectively, but i think this is a wrong way..
118331
Show that the system $x^{'}=x-y-x^{3}$,$y^{'}=x+y-y^{3}$ has a periodic solution.   I converted to polar:  $r r^{'}=x^{'}x+y^{'}y.$  Thus  $r r^{'}=x^{2}-x^{4}+y^{2}-y^{4}.$  Collecting squares:  $rr^{'}=x^{2}+y^{2}-(x^{2}+y^{2})^{2} +2x^{2}y^{2}.$  Substitute:  $r^{'}=r-r^{3}(1-(\sin^{2}(2\theta))/2)$  $r^{'}=r-r^{3}(1+\cos^{2}(2\theta))/2.$  Setting $r^{'}=0$ yields  $r^{2}=2/(1+\cos^{2}(2\theta).$  Thus $r_{max}= 2^{(1/2)}$ and $r_{min}=1.$  I suck at trig and this doesn't look right. Clearly $(0,0)$ is the only fixed point, so I have a bounded region. I just don't think I have found correct values for $r_{min}$ and $r_{max}$ to find my limit cycle.
39541
Let $A =  \mathbb{Z}[\sqrt{-2}]= \{a+b\sqrt{-2} \ : a, b \in \mathbb{Z}\}$. Show that the only natural number $x$ for which $x+\sqrt{-2}$ is a cube in $A$ is $x=5$.   So I have to show that there exists $c,d \in \mathbb{Z}$ such that $(x+\sqrt{-2})^3= c + d \sqrt{-2}$. From the last equation, I obtain $x(x^2-6) = c \in \mathbb{Z}$ and $3x^2-2= b \in \mathbb{Z}$, and here I blocked. Could I conclude from both equations? If not, is there exists another way to proceed (hint)?  Thanks!
38165
Suppose $\forall (a;b;c;d) \in \mathbb{Z}^4$ as $\gcd(a,b)=\gcd(c,d)=1$:  How can I prove that $\gcd(a,bc)=\gcd(a,c)$?  Also, how to prove that $\gcd(ac,bd) = (\gcd(a,d))*(\gcd(b,c))$?
62281
Question : Do there exist an infinite number of integer-solutions $(x,y,z)$ of $x^x\cdot y^y=z^z$ where $1\lt x\le y$ ?  Motivation : After struggling to find a solution, I've just got one solution, which is  $$(x,y,z)=(1679616, 2985984, 4478976).$$  In the following, I'm going to write how I got this solution.  Letting $d$ be the greatest common divisor of $x,y,z$, we can represent $$x=ad, y=bd, z=cd$$ where $a,b,c$ are coprimes with each other. Then, we get $$d^{a+b-c}\cdot a^a\cdot b^b=c^c.$$  In the following, let's consider without the condition $x\le y$. Here, I suppose  $$a=2^m, b=3^n, a+b-c=1.$$  (As a result, this supposition works.)  Then, we get $$d=\frac{c^c}{2^{ma}\cdot 3^{nb}}.$$  Hence, letting $c=2^k\cdot 3^l$, if  $$kc\ge ma=m\cdot 2^m, lc\ge nb=n\cdot 3^n,$$ then $d$ is an integer.  Since $(m,n)=(4,2)$ satisfies the above conditions, then we get $d=2^8\cdot 3^6=186624.$ Hence we can get  $$x=9d=2^8\cdot 3^8=1679616, y=16d=2^{12}\cdot 3^6=2985984, z=24d=2^{11}\cdot 3^7=4478976.$$ Note that here I interchanged $x$ and $y$.  P.S : I was surprised to get this solution because I got this almost by chance. So, I don't know the other solutions. If you have any helpful information, please teach me.
103936
This is some old Olympiad problem (I vaguely remember its Iranian )  I factorised it as  $ (x-y)(3x + 3y +1) = y^2 $  Then I fruitlessly tried to prove   $\gcd(x-y, 3x + 3y +1 ) = \gcd (6x+1,6y+1) = 1$  for proving that $x-y$ is a perfect square.  How do I continue my method or are there other ways to approach this intuitively?
47280
I am having trouble solving the diophantine equation given in the title. This is how far I came:  We can factor in $\mathbb{Z}[i]$ $y^2+4=x^3\Rightarrow (y+2i)(y-2i)=x^3$. I want to show now that their gcd is a unit so in $\{\pm 1,\pm i\}$. Suppose $d=u+vi$ is the gcd, then $u+vi\text{ | }2y$ and $u+vi\text{ | }4i$. Now taking the norm, we find $u^2+v^2\text{ | }4y^2$ and $u^2+v^2\text{ | }16$. From the second part we find that $d=u+vi\in\{\pm 1,\pm i, \pm 4,\pm 4i,\pm 1\pm i,\pm 2\pm 2i\}$. Obviously $\pm 4$ and $\pm 4i$ cannot divide $y\pm 2i$. I don't know how to show it for the others.   If we assume that $y+2i$ and $y-2i$ are coprime and using the equation $(y+2i)(y-2i)=x^3$ we find that $y-2i=\epsilon \alpha^3$ and $y+2i=\delta \beta^3$ where $\epsilon$ and $\delta$ are units. From here on it is straightforward to find what $\alpha$ and $\beta$ can be. My only problem is showing that $y-2i$ and $y+2i$ are coprime.
41594
I want to prove $gcd(a, a+2)$ = 1 or 2 for every integer $a$.   I know that (a, a+2) is equal to the least positive value of $ax + (a+2)y$, where $x, y \in \mathbb{Z}$. However, I'm a little stuck at this point. I'm thinking I should split apart the problem into two cases, one where $a$ is odd and one where $a$ is even? Any suggestions? Thanks.
14282
I was presented with the following problem:      Prove that there exist solutions to $x^2+y^2=z^n$ for all $n$, with $x,y,z, n \in \mathbb{N}$   I showed that by taking any Pythagorean triple $x^2+y^2=z^2$ and multiplying by $z^{2(n-1)}$ we get $(z^{n-1}x)^2+(z^{n-1}y)^2=(z^2)^n$, which allows me to generate solutions easily for any value of $n$. I managed to find several similar questions on this site, such as  concerning the specific case $n=3$. I notice that all of these questions take a similar approach and start with a Pythagorean triple and use it to generate general solutions.  Is there a way to prove the statement (or better yet provide solutions to the equation) without first relying on Pythagoras?