Solving a factorial equation

Question

Prove that the only solution to $k! +m! =n!$ is $k = 1, m = 1, n = 2$

How would you go about this? I can't seem to figure out where to start.

Answer 1

Clearly, $n>k,m$.

• $n>2\to 2n!=n!+n!>2\,k!+2\,m!$; there is no solution.

• $n=2\to1!+1!=2!$

Answer 2

Note that there are also other solutions, e.g. $m=k=0,n=2$.

Assuming now that we are looking for solutions $n,m,k\neq0$. Also note that $n\geq2$.

Next by assuming $k!\ge m!$ :
$$k!+m!\leq k!+k!=2k! \iff n! \leq 2k!$$ $$\iff \dfrac{n!}{k!} \leq 2,$$

from which the result follows. The final reasoning is left for you.

Answer 3

If you assume, wolog, $k\le m$ and you divide by $k!$ you get:

$1 + [(k+1).....m]= [(k+1).....m][(m+1)....n]$

Which is only possilbe if $[(k+1).....m] =1$.

Which is $k < m$ means $k=0$ and $m=1$.

......

Just finesse a bit allow for $k = m$ and divide allow "$[(k+1).....m]$" to be $1$ you get, if you divide by $k!=m!$:

$1 + 1= [(m+1)....n] = 2$

which means either $n = m+1=2$ or $2=(m+1)(m+2)=(m+1)n$

which means $m=k=1$ and $n=2$ or $m=k=0$ and $n=2$.

So only solutions are $0! + 0! = 2=2!$

$0! + 1! = 2!$

or

$1! + 1! = 2!$.