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October 25th, 2009, 12:54 PM
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| | 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 . | 
October 26th, 2009, 12:35 PM
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| | Quote:
Originally Posted by miss blue 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
such that p,q are prime numbers
g.c.d(a,b) the product of the similar primes in a and b say
x is integer number, there exist such x since g.c.d is square
y is integer, there exist such y since l.c.m is square
note that  in other word  and  are relatively primes
g.c.d(a,b) ,  ,  all of them are squares
say
s,t are integers now
so a,b are squares
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November 7th, 2009, 06:45 AM
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| | thx alot Amer
but plz if u have a selution for the athors be generous
ant kareem wa n7n nastahal
thx | 
November 8th, 2009, 08:20 AM
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| | Quote:
Originally Posted by miss blue 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
or
m>0 , is there any conditions on a
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November 8th, 2009, 08:42 AM
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| | yessss sorry
i copied it wrong
a>1
thx | 
November 8th, 2009, 09:57 AM
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note that
r is the reminder such that
now
but
r is the reminder
to find r sub a=1 you will have
the proof is finished
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Last edited by Amer; November 9th, 2009 at 06:36 AM.
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