Linked Questions

0
votes
0answers
141 views

is $\sqrt{n!}\notin\mathbb{Z}$ for $n>1$ true? [duplicate]

Is it true that $\sqrt{n!}\notin\mathbb{Z}$ for $n>1$? This is what I did: By induction: for $n=2$ its trivial ($\sqrt{2}$ is irrational). Suppose its true for some $n\in\mathbb{N}$, then the ...
0
votes
0answers
29 views

Is there any number which is a perfect square of some number as well as factorial of some number. [duplicate]

Is there any number which is a perfect square of some number as well as factorial of some number? That is, let x be such a number, then x can be expressed as ...
27
votes
2answers
4k views

Is $\sqrt{n!}$ a natural number?

I'm new here (on Mathematics Stack Exchange). Also, I'm a 10th grade student not a math expert. So, my question is whether, $$\sqrt {n!} $$ comes in the set of the Natural Numbers. There were ...
103
votes
1answer
5k views

$n!$ is never a perfect square if $n\geq2$. Is there a proof of this that doesn't use Chebyshev's theorem?

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 $...
4
votes
3answers
1k views

Find all values of $n$ greater than or equal to 1 for which $n! + (n + 1)! + (n + 2)!$ is equal to a perfect square.

Not sure where to get started on this on. I started listing numbers for n starting at 1 but the numbers get very big very fast and I cannot find a pattern. Is there a better way of doing this or ...
7
votes
3answers
221 views

Find all natural numbers $n > 1$ and $m > 1$ such that $1!3!5!\cdots(2n - 1)! = m!$

Find all natural numbers $n > 1$ and $m > 1$ such that $1!3!5!\cdots(2n - 1)! = m!$ I have been thinking about coming up with some inequalities which would narrow the possible range of pairs $(...
4
votes
1answer
731 views

Factorials and their perfect squares

How many positive factorials are also perfect Squares. So for example $1!=1=1^2$. How many others exist other than 1? Is there any way to prove this?
0
votes
1answer
443 views

Prove: $\lfloor \sqrt{n!}\rfloor\nmid n!$ [closed]

$n$ is an integer greater than $7$. How does one go about proving that $\lfloor \sqrt{n!}\rfloor\nmid n!$.
1
vote
2answers
131 views

Prove that $L_1$ = $\{1^m :$ m is not a perfect square$\}$ is not regular using Pumping Lemma

I'm trying to prove that the Language $L_1$ = $\{1^m :$ m is not a perfect square$\}$ is not regular. I proved before that L = $\{1^m :$ m is a perfect square$\}$ is not regular, I thought that I ...
0
votes
2answers
46 views

Occurrence (parity) of primes in $n!$

It is known that $n!$ can not be a perfect square for $n\geq 1$. This means that in the prime decomposition of $n!$, one of the prime occurs odd number of times. This leads to following two questions: ...