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Math Help - powers and factorials

  1. #1
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    powers and factorials

    let p, r_1 ,r_2,s_1,s_2 be positive integers with
    p>1 ;  r_1 < r_2  and s_1<s_2 then

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

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

    is it correct ?

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

    plz help.
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  2. #2
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    Quote Originally Posted by thippli View Post
    let p, r_1 ,r_2,s_1,s_2 be positive integers with
    p>1 ;  r_1 < r_2  and s_1<s_2 then

    (1) p^{r_1}+ p^{r_2}= p^{s_1}+  p^{s_2}
    if and only if r_i = s_i, i=1,2 .
    Note that p^{r_1}+p^{r_2}=p^{r_1}(1+p^{r_2-r_1}), and 1+p^{r_2-r_1} is not divisible by p (except if r_1=r_2 and p\neq 2, but remember r_1<r_2). As a consequence, r_1 is the largest number m such p^m divides p^{r_1}+p^{r_2}. Can you see why this answers your question ?


    (2) r_1r_1!+ r_2 r_2!= s_1 s_1!+ s_2s_2!
    if and only if r_i = s_i, i=1,2.
    You can solve this by the same method like above. Remember that r_2!=r_1! \times (r_1+1)(r_1+2)\cdots (r_2-1)r_2.
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  3. #3
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    Thank you

    First answer is very clear. Thank you very much !

    But I think the second won't work , because

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

    Here r_1 may divide r_2. In this case how can we say r_1r_1! is the largest integer which divides
     <br />
1+ \frac{r_2} {r_1} \frac{r_2!}{r_1!}<br />
?
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  4. #4
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    Quote Originally Posted by thippli View Post
    First answer is very clear. Thank you very much !

    But I think the second won't work , because

    r_1r_1 ! + r_2r_2! = r_1 r_1!(1+ c).

    Here r_1 may divide r_2. In this case how can we say r_1r_1! is the largest integer which divides
     <br />
1+ \frac{r_2} {r_1} \frac{r_2!}{r_1!}<br />
?
    You don't even know if \frac{r_2} {r_1} \frac{r_2!}{r_1!} is an integer, so this is not the right factorization. What works however is:
    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 r_1+1 does not divide the second factor, so that r_1 is the largest n such that n! divides r_1r_1!+r_2r_2!.
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