Results 1 to 6 of 6

Math Help - any help plz

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    7

    any help plz

    hi every one
    can any buby help me with these questions?

    1. If a,b are in N*, if gratest common div.(a,b) & smallest com. mult.(a,b) are squares, show that a,b are squares.

    2. Show that for every m>0 , a?1 we have

    ( a^m -1 / a-1 , a-1 ) = ( a-1 , m ).

    3. show that any two terms in the sequance: 2+1, 2^2+1, 2^4+1, .. , 2^2n+1 are coprime .
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by miss blue View Post
    hi every one
    can any buby help me with these questions?

    1. If a,b are in N*, if gratest common div.(a,b) & smallest com. mult.(a,b) are squares, show that a,b are squares.

    2. Show that for every m>0 , a?1 we have

    ( a^m -1 / a-1 , a-1 ) = ( a-1 , m ).

    3. show that any two terms in the sequance: 2+1, 2^2+1, 2^4+1, .. , 2^2n+1 are coprime .

    let

    a=p_1\times p_2 \times ... \times p_n

    b=q_1\times q_2 \times ...\times q_m

    such that p,q are prime numbers
    g.c.d(a,b) the product of the similar primes in a and b say

    g.c.d(a,b)=p_1\times p_2 \times ... \times p_i=q_1\times q_2 \times ... \times q_i=x^2

    x is integer number, there exist such x since g.c.d is square

    l.c.m(a,b)=g.c.d(a,b)\times (p_{i+1}\times ...\times p_n)(q_{i+1}\times ...\times q_m)=y^2

    y is integer, there exist such y since l.c.m is square


    note that
    \left(\frac{a}{g.c.d(a,b)},\frac{b}{g.c.d(a,b)}\ri  ght)=1 in other word

    (p_{i+1}\times ...\times p_n) and (q_{i+1}\times...\times q_m) are relatively primes


    g.c.d(a,b)\times (p_{i+1}\times ...\times p_n)(q_{i+1}\times ...\times q_m)=y^2

    g.c.d(a,b) , (p_{i+1}\times ...\times p_n) , (q_{i+1}\times ...\times q_m) all of them are squares

    say
    (p_{i+1}\times ...\times p_n)=t^2

    q_{i+1}\times ...\times q_m)=s^2

    s,t are integers now

    a=x^2(t^2)

    b=x^2(s^2)

    so a,b are squares
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    7
    thx alot Amer

    but plz if u have a selution for the athors be generous
    ant kareem wa n7n nastahal

    thx
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by miss blue View Post
    hi every one
    can any buby help me with these questions?

    1. If a,b are in N*, if gratest common div.(a,b) & smallest com. mult.(a,b) are squares, show that a,b are squares.

    2. Show that for every m>0 , a?1 we have

    ( a^m -1 / a-1 , a-1 ) = ( a-1 , m ).

    3. show that any two terms in the sequance: 2+1, 2^2+1, 2^4+1, .. , 2^2n+1 are coprime .
    in the second question you want to prove that

    \left( \frac{a^{m-1}}{a-1} , a-1 \right) = (a-1 , m )

    or

    \left( \frac{a^m -1 }{a-1} , a-1 \right) = ( a-1 , m )

    m>0 , is there any conditions on a
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2009
    Posts
    7
    yessss sorry
    i copied it wrong

    a>1

    thx
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    \left(\frac{a^m-1}{a-1} , a-1 \right) = (a-1 , m )

    note that

    (a,b)=(r,b)

    r is the reminder such that a=xb+r

    now

    \frac{a^m-1}{a-1} = a^{m-1} + a^{m-2} + ...+ 1

    but

    ( a^{m-1} + + ...+ 1 , a-1 ) = (a-1 , r )

    r is the reminder

    a^{m-1} + + ...+ 1 = (a-1)(s) + r

    to find r sub a=1 you will have

    1+1+...+1 = 0(s) + r \Rightarrow r=m

    the proof is finished
    Last edited by Amer; November 9th 2009 at 04:36 AM.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum