Find all prime number $p$ for which $p! + p$ in a perfect square.  Of course $p=2$ and $p=3$ are solutions.  I think there are not other solutions, but I can’t prove it!
72873
Find all prime numbers $p$ such that $p!+p$ is a perfect square.  I think the only ones are when $p!+p=p^2$, i.e. $p\in \{2,3\}$.  Any ideas at all?
112475
I am trying to prove that ($\forall \ n\in\mathbb{N}$) there exists a prime number $q$ such that $n &lt; q \le 1 + n!$  I have made a graph with $n=0$ through $n=10$ and found solutions to all of them looking for a pattern and I see that $n!$ gets enormous fast and it becomes quite obvious that there is a prime number in between them.    I have considered trying to prove by contradiction that $q$ does not exist on that interval, but I don't know where to go from that statement.  Could anybody help me figure it out?  I have been staring at it for hours and I can't figure out where to go.  Thank you.
107493
Problem Given a number n, $2 \leq n \leq 2^{63}$. $n$ could be prime itself. Find the a prime $p$ that is closet to $n$.  Using the fact that for all prime $p$, $p &gt; 2$, $p$ is odd and $p$ is of the form $6k + 1$ or $6k + 5$, I wrote a for-loop checking from $n - 1$ to 2 to check if that number is prime. So instead of checking for all numbers I need to check for every odd of the two form $p \equiv 1 \pmod{k}$, and $p \equiv 5 \pmod{k}$. However, I wonder if there is a faster algorithm to solve this problem? i.e. some constraints that can restrict the range of numbers need to be checked? Any idea would be greatly appreciated.
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Problem: If $p\geq 5$ is a prime number, show that $p^2+2$ is composite.  Remarks: Now if one observes that $p$ takes the forms $6k+1$ and $6k+5$, the problem is resolved quite easily. However, if one were to choose other forms say $4k+1$ and $4k+3$ then $p^2$ would be of the form $4p+1$ which would yeild the general form $4p+3$ on addition with $2$. Obviously this does not lead to any concrete conclusion. I was wondering whether there is any specific procedure involved in finding the right quotients (For eg. $6$ and $4$) or is this just a random problem solving trick. Moreover, if anyone could provide some intuition as to why $6$ works would be much appreciated.
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Can $n!$ be a perfect square when $n$ is an integer greater than $1$?  Clearly, when $n$ is prime, $n!$ is not a perfect square because the exponent of $n$ in $n!$ is $1$.  The same goes when $n-1$ is prime, by considering the exponent of $n-1$. What is the answer for a general value of $n$?  (And is it possible, to prove without Bertrand's postulate. Because Bertrands postulate is quite a strong result.)
104255
can someone please describe me why only the perfect square has odd number of factors.why does other number not has odd numbers of factors? I understand it but don't find any mathmetical proof.Please help me
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If $n\geq2$, then $n!$ is not a perfect square. The proof of this follows easily from Chebyshev's theorem, which states that for any positive integer $n$ there exists a prime strictly between $n$ and $2n-2$. A proof can be found .  Two weeks and four days ago, one of my classmates told me that it's possible to prove that $n!$ is never a perfect square for $n\geq2$ without using Chebyshev's theorem. I've been trying since that day to prove it in that manner, but the closest I've gotten is, through the use of the prime number theorem, showing that there exists a natural number $N$ such that if $n\geq N$, $n!$ is not a perfect square. This isn't very close at all. I've tried numerous strategies over the past weeks and am now trying to use the Sylow theorems on $S_n$ to somehow show that $|S_n|$ can't be square (I haven't made any progress).  Was my classmate messing with me, or is there really a way to prove this result without Chebyshev's Theorem? If it is possible, can someone point me in the right direction for a proof?  Thanks!
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Assume: $p$ is a prime that satisfies $p \equiv 3 \pmod 4$  Show: $x^{2} \equiv -1 \pmod p$ has no solutions $\forall x \in \mathbb{Z}$.  I know this problem has something to do with Fermat's Little Theorem, that $a^{p-1} \equiv 1\pmod p$. I tried to do a proof by contradiction, assuming the conclusion and showing some contradiction but just ran into a wall. Any help would be greatly appreciated.
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I came to this conclusion: the primes that can't be written as the sum of two squares (lets use$p\prime$ to denote) can be written as ${p}\prime \equiv 3 \;(\bmod\; 4)$.  I need a simple proof to show this and am stuck. I would appreciate guidance!
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I'd like to have my proof verified and if possible, to see other solutions that are interesting.  Proof: Suppose $n(n+1)$ is a square. Then we write $$n(n+1) = \prod_{p} p^{c(p)}$$ where $c(p) = a(p) + b(p)$ are such that \begin{align*} n &amp;= \prod_p p^{a(p)} \\ n+1 &amp;= \prod_p p^{b(p)} \end{align*} Now, by our hypothesis, $c(p)$ is even for all primes $p$. As $(n,n+1)=1$ for all $n$, it must be that $a(p)$ and $b(p)$ are even for all primes $p$ and moreover, $a(p) = 0$ whenever $b(p)&gt;0$ and reversely.  This indicates that both $n$ and $n+1$ are squares. This is impossible as there are no consecutive squares in the natural numbers.
113249
How many positive integer $n$ are there  such that $2n+1$ , $3n+1$ are both perfect squares ?   $n=40$ is a solution . Is this the only solution ? Is it possible to tell whether finitely many or infinitely many solutions exist ?
91749
$x^2+1=0$ in $\mathbb{Z}_7$   By trying each number, I see that there is no solution, is this correct?  And could you help me with a more direct solution, since this method is not going to work for $\mathbb{Z}_p$ when $p$ is a large prime number .  Thank you very much!
100178
Find p is the prime number which $\dfrac{p+1}{2}$ and $\dfrac{p^2+1}{2}$ both are square number.  I do not know how to use "p is prime" assumption given. I just know  $p=7$ is satisfied.  If $\dfrac{p+1}{2}=X^2$ and $\dfrac{p^2+1}{2}=Y^2$ then $(X^2;X^2-1;Y)$ is Pythagorean triple
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Let $p$ be a prime number. Show that the congruence $x^{p-1} \equiv 1 \pmod {p^2}$ has precisely $p-1$ solutions modulo $p^2$  I know the statment above is true for modulo $p$, but how is this imply to modulo $p^2$
107122
Given $p$ is prime number, find all pairs of positive integers $m,n$ such that $\gcd(m,n)=1$ and  $$\frac{p+m}{p+n}-\frac{m}{n}=\frac{1}{p^{2}}$$  We can write this equation as: $$\frac{p(n-m)}{n(p+n)} = \frac{1}{p^2}$$    Since we have positive RHS and $n(p+n)&gt;0$, we also have: $$p(n-m)&gt;0 \implies n&gt;m$$ I found that $n&gt;m$, but I have no idea to complete. Please give me some ideas, thank you very much!!
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Are there any positive $n$ for which $ n^4+n^3+n^2+n+1$ is a perfect square?  I tried to simplify   \begin{align*} n^4+n^3+n^2+n+1 &amp;= n^2(n^2+1)+n(n^2+1)+1\\  &amp;= (n^2+n)(n^2+1)+1 \\   &amp;= n(n+1)(n^2+1)+1 \end{align*} Then I assumed that the above expression is a square; then  $$ n(n+1)(n^2+1)+1 = k^2$$  $$  \begin{align*} n(n+1)(n^2+1) &amp;= (k^2-1) \\    &amp;= (k+1)(k-1) \end{align*} $$   Then trying to reason with prime factors, but cannot find a concrete proof yet.
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Find all integers $x$ such that $x^2+3x+24$ is a perfect square.    My attempt:    $x^2+3x+24=k^2$ $3(x+8)=(k+x)(k-x)$    Now, do I find solution treating cases? But that doesn't seem very easy. Please help.