How can I solve this Diophantine equation: $$(n-1)!+1=n^m$$ with $n,m$ positive integers?  From Wilson's theorem we can note that $n$ is a prime number. I proved to rewriting the equation as:$$(n-2)!=n^{m-1}+n^{m-2}+....+1$$ but in vain. I proved to solve also through the theorem LTE bur I analysed only some cases without obtain a general solution.
14962
How to find all positive integers $n$ such that $(n-1)!+1$ can be written as $n^k$ for some positive integer $k$?
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.
1749
Suppose that $p$ is a prime. Suppose further that $h$ and $k$ are non-negative integers such that $h + k = p − 1$.   I want to prove that $h!k! + (−1)^h \equiv 0 \pmod{p}$  My first thought is that by Wilson's theorem, $(p-1)! \equiv -1 \pmod{p}$, and $h!k!$ divides $(p-1)!$ (definition of a binomial). Where would I go from here?
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?
25053
$\forall m,n\in\mathbb Z$ , $m\ge1$ and $n\ge1$ how to prove that  $$\frac{(mn)!}{m!(n!)^m}$$ is an integer?  Thanks in advance.
122670
... namely that if $n &gt; 1$ and $(n − 1)!\equiv−1\pmod{n}$, then $n$ is prime.  This is for a number theory class I'm in at Penn State. My idea is to follow accordingly, but I can't get it concretely..  Reason by contradiction, suppose that $n$ is not prime.   Then $n = ab$, where $1 &lt; a, b &lt; n$  In particular $(n-1)! = ai$ for some $i\in\Bbb{Z}$. $\\$ If $(n-1)! = nj - 1$ for some $j\in\Bbb{Z}$ then it follows that a probably doesn't divide $-1$?
90185
Well, I've confused when trying to solve this equation can anybody help me :  $n^2 \equiv 0 \pmod{584}$  I tried to factorize the $584$ i got $584=2^3\times73$.  so $n^2$ has to be divisible by $2^3$ and $73$ in this same time.  here i get stuck.
74062
Given a linear Diophantine equation, how can I count the number of positive solutions?  More specifically, I am interested in the number of positive solutions for the following linear Diophantine equation:  $3w + 2x + y + z = 47$  Update: I am only interested in non-zero solutions.
61352
If $n$ be a positive integer greater than $1$, then prove that $$2^n&gt;1+n\sqrt{2^{n-1}}$$    I found this problem under $AM \ge GM$ chapter. Help me to solve this problem using  $AM \ge GM$. Thanks in advance.
71339
$$p+1=2n^2$$$$p^2+1=2m^2$$ Find positive integers $m,n$ and prime $p$ satisfying the above two equations.   What would people commonly do?  Subtracting both the equations.  You get:   $$p(p-1)=2(m-n)(m+n)$$  If you notice carefully, the above equation has infinitely many solutions, and this doesn't work because the generality of the equation is lost when you subtract both of them.  I noticed that $p \equiv -1 \pmod 8$. $p=7$ works fine. I don't see any other solution, but not able to contradict of existence.
44490
I realize that there is a similar post to this, but that post included a hint which we were not given. Also regarding that hint, I'm just wondering how someone could find it out for themselves. Here is the related post:  I know that $F_n = 2^{2^n} + 1$ and therefore $\,2^{2^n} \equiv -1 \pmod{F_n}$. But I don't intuitively know the next step to take to show that $\ 2^{F_n}\equiv 2\pmod{\!F_n},\,$ i.e. that  $$2^{2^{2^n} + 1} \equiv 2 \pmod{2^{2^n}+1}$$ So in essence, I would like to know how to solve it without knowing the hint. Thank you.
6444
One observes that  \begin{equation*} 4!+1 =25=5^{2},~5!+1=121=11^{2}  \end{equation*} is a perfect square. Similarly for $n=7$ also we see that $n!+1$ is a perfect square. So one can ask the truth of this question:   Is $n!+1$ a perfect square for infinitely many $n$? If yes, then how to prove.
5493
I need to prove the following, but I am not able to do it. This is not homework, nor something related to research, but rather something that came up in preparation for an exam.     If $n = 1 + m$, where $m$ is the product of four consecutive positive   integers, prove that $n$ is a perfect square.   Now since $m = p(p+1)(p+2)(p+3)$;     $p = 0, n = 1$ - Perfect Square      $p = 1, n = 25$ - Perfect Square      $p = 2, n = 121$ - Perfect Square   Is there any way to prove the above without induction? My approach was to expand $m = p(p+1)(p+2)(p+3)$ into a 4th degree equation, and then try proving that $n = m + 1$ is a perfect square, but I wasn't able to do it. Any idea if it is possible?
30096
Verify that $4(29!)+5!$ is divisible by $31$  I know I have to use Wilsons theorem: $(p-1)!=-1\pmod p$ but I'm not really sure how to apply this theorem. Step by step explanation please? Thank you!
30555
What are the integer solutions to the diophantine equation   $$n^2 + 1 = 2 \times 5 ^m? $$  We have $(n,m) = (3,1), (7, 2) $ as solutions. Are there any more?    This seems like it would be a well known diophantine equation, but I can't seem to find any information about it.
104049
Prove that there does not exist an integer solution for the diophantine equation $x^5 - y^2 = 4$.   It's obvious that $x$ and $y$ are of the same parity. We can also claim that if $x$ is odd, then it is $1 \pmod 4$. Also, if $x$ and $y$ are even, then $y \equiv 2 \pmod 4\text{ since } x^5 \text{ is a multiple of } 32$ and if $y$ were a multiple of $4$ then it would be at a distance of at least $16$ from $x^5$ or be equal to it. These are my observations.  How should I proceed with the proof? Please give hints.
60238
For any integer $a,b$ let $N_{a,b}$ denote the number of positive integer $x&lt;1000$ satisfying $x= a \mod 27$ and $x=b \mod 37$. Then which of them is correct         there exist $a,b$ such that $N_{a,b} =0$    for all $a,b$, $N_{a,b}=1$   for all $a,b$, $N_{a,b}&gt;1$   there exist $a,b$ such that $N_{a,b}$ and  there exist $a,b$ such that $N_{a,b}=2$.      Since we know that by Chinese Remainder Theorem if $m_1, m_2, .., m_k$ are pairwise relatively prime positive integers, and if $a_1, a_2, .., a_k$ are any integers, then the  simultaneous congruences $x ≡ a_1 (mod\;m_1), x ≡ a_2 (mod\;m_2), ..., x ≡ a_k (mod\;m_k)$ have a solution, and the solution is unique modulo $m$, where  $m = m_1m_2⋅⋅⋅m_k$. Then how we will approach by acoording to this theorem. So please someone help me and explain it.