# powers and factorials

• Jan 22nd 2009, 10:52 PM
thippli
powers and factorials
let $\displaystyle p, r_1 ,r_2,s_1,s_2$ be positive integers with
$\displaystyle p>1 ; r_1 < r_2$ and $\displaystyle s_1<s_2$ then

(1) $\displaystyle p^{r_1}+ p^{r_2}= p^{s_1}+ p^{s_2}$
if and only if $\displaystyle r_i = s_i$, $\displaystyle i=1,2$ .

(2) $\displaystyle r_1r_1!+ r_2 r_2!= s_1 s_1!+ s_2s_2!$
if and only if $\displaystyle r_i = s_i$, $\displaystyle i=1,2$.

is it correct ?

the one way in each case is obvious. the other way...?

plz help.
• Jan 24th 2009, 03:22 AM
Laurent
Quote:

Originally Posted by thippli
let $\displaystyle p, r_1 ,r_2,s_1,s_2$ be positive integers with
$\displaystyle p>1 ; r_1 < r_2$ and $\displaystyle s_1<s_2$ then

(1) $\displaystyle p^{r_1}+ p^{r_2}= p^{s_1}+ p^{s_2}$
if and only if $\displaystyle r_i = s_i$, $\displaystyle i=1,2$ .

Note that $\displaystyle p^{r_1}+p^{r_2}=p^{r_1}(1+p^{r_2-r_1})$, and $\displaystyle 1+p^{r_2-r_1}$ is not divisible by $\displaystyle p$ (except if $\displaystyle r_1=r_2$ and $\displaystyle p\neq 2$, but remember $\displaystyle r_1<r_2$). As a consequence, $\displaystyle r_1$ is the largest number $\displaystyle m$ such $\displaystyle p^m$ divides $\displaystyle p^{r_1}+p^{r_2}$. Can you see why this answers your question ?

Quote:

(2) $\displaystyle r_1r_1!+ r_2 r_2!= s_1 s_1!+ s_2s_2!$
if and only if $\displaystyle r_i = s_i$, $\displaystyle i=1,2$.
You can solve this by the same method like above. Remember that $\displaystyle r_2!=r_1! \times (r_1+1)(r_1+2)\cdots (r_2-1)r_2$.
• Feb 2nd 2009, 03:21 AM
thippli
Thank you
First answer is very clear. Thank you very much !

But I think the second won't work , because

$\displaystyle r_1r_1 ! + r_2r_2! = r_1 r_1!(1+ \frac{r_2} {r_1} \frac{r_2!}{r_1!})$.

Here $\displaystyle r_1$ may divide $\displaystyle r_2$. In this case how can we say $\displaystyle r_1r_1!$ is the largest integer which divides
$\displaystyle 1+ \frac{r_2} {r_1} \frac{r_2!}{r_1!}$ ?
• Feb 3rd 2009, 03:25 AM
Laurent
Quote:

Originally Posted by thippli
First answer is very clear. Thank you very much !

But I think the second won't work , because

$\displaystyle r_1r_1 ! + r_2r_2! = r_1 r_1!(1+ c)$.

Here $\displaystyle r_1$ may divide $\displaystyle r_2$. In this case how can we say $\displaystyle r_1r_1!$ is the largest integer which divides
$\displaystyle 1+ \frac{r_2} {r_1} \frac{r_2!}{r_1!}$ ?

You don't even know if $\displaystyle \frac{r_2} {r_1} \frac{r_2!}{r_1!}$ is an integer, so this is not the right factorization. What works however is:
$\displaystyle r_1 r_1!+r_2 r_2!=r_1!\left(r_1+r_2\frac{r_2!}{r_1!}\right)=r_1 !\left(r_1+r_2(r_1+1)(r_1+2)\cdots r_2\right).$
Notice indeed that $\displaystyle r_1+1$ does not divide the second factor, so that $\displaystyle r_1$ is the largest $\displaystyle n$ such that $\displaystyle n!$ divides $\displaystyle r_1r_1!+r_2r_2!$.