I am attempting to find solutions to $x^y = y^x$. I have been trying to find solutions to this equation for quite some time now, and using all the math (and common sense) I know, I arrived at one possible non trivial solution, $x = 4, \;y =2.$ Are there other solutions to this equation and if there are how do I find them?
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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 $x^y = y^x$?
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How to find all solutions for Pell's equation $x^2 - Dy^2 = -1$ after the first $x_0$ and $y_0$?   for example if we have $x^2 - 2 y^2 = -1$ then the smallest integer solution for $(x,y) = (1,1)$  How to build on this to get the rest of the solutions ?
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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 :)
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Exercise $12$ in Section $1.6$ of Nathanson's : Methods in Number Theory book has the following question.         When is the sum of a geometric progression equal to a power? Equivalently, what are the solutions of the exponential diophantine equation $$1+x+x^{2}+ \cdots +x^{m} = y^{n} \qquad \cdots \ (1)$$ in integers $x,m,n,y$ greater than $2$? Check that    \begin{align*} 1 + 3 + 3^{2} + 3^{3} + 3^{4} &amp; = 11^{2}, \\\ 1 + 7 + 7^{2} + 7^{3} &amp;= 20^{2}, \\\ 1 + 18 +18^{2} &amp;= 7^{3}. \end{align*}   These are the only known solutions of $(1)$.      The  doesn't reveal much about the above question. My question here would be to ask the following:   Are there any other known solutions to the above equation. Can we conjecture that this equation can have only finitely many solutions?   Added: Alright. I had posted this question on  some time after I had posed here. This user by name  had actually given me a link which tells more about this particular question. Here is the link:
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My question is this:      $y^2 = \frac{x^5 - 1}{x-1}$ &amp; $x,y \in \mathbb{Z}$     Source: BdMO 2016 Regionals Set 1 Question 3.   I think this is the exact same question. Actually I didn't get the solution in the link. Please don't mark as duplicate.  I tried solving this same problem by writing code; there seems to be $6$ solution pairs to this equation. How do I get the $6$ solutions mathematically? I indeed need a full solution. Please help.
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Find the number of solutions to the equation $$2011^x+2012^x+2013^x-2014^x=0$$  The answer seems to be zero, but I have no idea why. Please avoid considering complex solutions and other scary things.
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Find all real solutions for $x$ in $2(2^x−1)x^2+(2^{x^2}−2)x=2^{x+1}−2$.   I started off by dividing $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$ by $2$, and I got $(2^x-1)x^2 + (2^{x^2-1}-1)x = 2^x -1$. I tried dividing by $2^x-1$ on both sides which would give me a simple quadratic to solve for $x$, but I don't know how to simplify $\frac{(2^{x^2}-1)}{(2^x-1)}$. Am I missing a simple trick here, or am I on a completely wrong path to solving this?
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I am assigned to find solutions to the Pell-type equation $x^2-2y^2=-1$. So far, I only have $(7,5), (41,29)$ and $(239,169)$.  My question is, is there a general formula to find all its solutions? Are there infinitely many solutions? If it's possible to write Mathematica code that would find its solutions, what would the code look like?  Thanks a lot.
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I would like to know any way of solving the diophantine equation $x^4+y^4=2z^2$. Or ideas that seem worth trying out.  By solving I mean fining all solutions and proving there are no more.  Keith Conrad showed how to reduce this equation to a different one which was solved by Fermat in his notes about Fermat descent, other than I have no ideas how to solve it. I tried to do descent on it directly but that seems completely impossible so I am interested in other techniques. Thanks very much.
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Given that   $x^x = y$; and  given some value for $y$   is there a way to expressly solve that equation for $x$?
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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.
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Find all integer solutions of the equation $a^2+ b^2 + c^2=a^2 b^2$.  This is one of the questions we presented in one session to contest preparation PUTNAM. It turns out that I can't get from the problem. Could someone just give me a hint? (Please, don't give me the answer. Simply, an argument that can help me advance in the problem or theorem might suffice.)
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I am trying to find the particular solution to the $y^{(4)} -2y'' +y = xe^x $ and currently am misunderstanding what to do.   My steps:   the polynomial operator concerned is $p(s)= s^4 -2s^2 + 1 $ which  0 at 1:  $p(1) = 0 $  so now i know that the solution will be something like:  $y_p = (Ax^3 + Bx^2)e^x$  where the parentheses show that it is a linear operator on the last coefficient.   so far I think that my method is to differentiate this four times and collect up the terms of each $y_p$ according to the original equation and see what A and B are equal to.   I assume this is okay to do with linear operators?   $y_p' = (Ax^3 + Bx^2)e^x + (3Ax^2 + 2Bx)e^x$  $y_p'' = (Ax^3 +(6A + B)x^2 + (4B + 6A)x + 2B)e^x$  can I collect terms from different linear operators like i did with $y_p''$   and I assume that to find the particular solution I must find $y_p''''$ as it is in the original equation
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The one of solution of $x^4 - 2y^2 = -1$ is $x = 1$ and $y = 1$. However, the solution $(1, 1)$  of $x^4 - 2y^2 = 1$ is failed. We know $x = 1$ and $y = 1$ is small integers and we can check by trail method. In case more solutions are existing or not how to check? What are the solutions of $x^4 - 2y^2 = 1$?
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Solve the equation $28^x = 19^y+87^z$, where $x,y,z$ are integers.   I am confused how to find solutions for all integers and not just positive ones. How should we do that?
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As far as I know, we are searching solutions in set of positive integers for $x^n + y^n = z^n$  for $n &gt; 2$. There are many proofs are stated that Fermat is true and there is no solutions for this equations. Now my question is, can we have solutions in complex numbers field? If yes, how to find them.   Thanks in advance.
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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 $4k+1$.