5
votes
1answer
185 views

Finding every triplet $(n,a,b)$ such that $n!=2^a-2^b$

Question : Let $n,a,b$ be positive integers. Are there infinitely many triplets $(n,a,b)$ which satisfy the following equality?$$n!=2^a-2^b$$ If Yes, then how can we prove that? If No, then how ...
1
vote
0answers
66 views

When is $n!+1$ a square? [duplicate]

I'm looking for the solutions $(n,m)$ of the equation $n!+1=m^2$. I have calculated the values of $\sqrt{n!+1}$ for $n \le $ and found only the solutions $(4,5)$, $(5,11)$ and $(7,71)$. Are these ...
3
votes
2answers
190 views

Factorials and Arithmetic Progression.

Are there sets of factorials $(a_1!,a_2!,a_3!,\dots,a_n!)$, such that they exist in Arithmetic progression. $n$ is a natural number I don't see any such examples(Except for $n=2$). And I don't see ...
7
votes
5answers
377 views

Find all solutions of the equation $x! + y! = z!$ [duplicate]

Not sure where to start with this one. Do we look at two cases where $x<y$ and where $x>y$ and then show that the smaller number will have the same values of the greater? What do you think?
1
vote
3answers
181 views

Number of solutions for $\frac{1}{X} + \frac{1}{Y} = \frac{1}{N!}$ where $1 \leq N \leq 10^6$

Note: this is a programming challenge at this site For this equation $$\frac{1}{X} + \frac{1}{Y} = \frac{1}{N!}\quad ( N \text{ factorial} ),$$ find the number of positive integral solutions for ...
19
votes
1answer
476 views

Integer solutions of $x! = y! + z!$

There was an interesting problem asked about triples $(x,y,z)$ which are solutions of $$x! = y! + z!.$$ Here $(2,1,1)$ is a solution because $2! = 1! + 1!$, as are $(2,1,0)$ and $(2,0,1)$. Now I ...
8
votes
1answer
280 views

Solutions of $p!q! = r!$

The title says it all, more or less. Obviously, there are infinitely many "trivial" integral solutions of the form $p=n, q=(n!-1), r= n!$. How many non-trivial solutions are there? I came across this ...
10
votes
2answers
316 views

Finding all the numbers that fit $x! + y! = z!$

I have the formula $x! + y! = z!$ and I'm looking for positive integers that make it true. Upon inspection it seems that x = y = 1 and z = 2 is the only solution. The problem is how to show it. ...
3
votes
3answers
492 views

Finding all positive integer solutions to $(x!)(y!) = x!+y!+z!$

The equation is $(x!)(y!) = x!+y!+z! $ where $x,y,z$ are natural numbers. How to find out them all?