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April 23rd, 2012, 04:31 AM   #1
 
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how to prove that if a=b(mod n) then ....

hey i'm not really sure how i would go about writing out this answer..

"prove that if a = b (mod n) then a^2 = b^2 (mod n) "
as in if a and b are congruent to modulo n .. then a squared and b squared are congruent to modulo n ...

thanks munchiez
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April 23rd, 2012, 04:42 AM   #2
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Re: how to prove that if a=b(mod n) then ....

By your problem, a isnt divisible by b.
let a^2 is divisible by b^2. Then (a/b)^2 = k where k is a integer. If k is a nonrootable number the we have nothing to do, it contradicts. If k is a rootable number, then a/b = p where p is an integer. By our primary assumption, it is impossible. So a^2 isnt divisible by b^2.
now by you assumption, a = b + nc where c is any integer. Then squaring both sides, it becomes
a^2 = b^2 + n(2bc + nc^2) so, a^2 = b^2 (mod n)
Q. E. D
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April 23rd, 2012, 04:51 AM   #3
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Re: how to prove that if a=b(mod n) then ....

Quote:
Originally Posted by mathbalarka
By your problem, a isnt divisible by b.
That doesn't follow. (a, b, n) = (2, 2, 2) is possible, and in that example a is divisible by b.
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April 23rd, 2012, 04:55 AM   #4
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Re: how to prove that if a=b(mod n) then ....

Yes, but what about my next solution,
Quote:
now by you assumption, a = b + nc where c is any integer. Then squaring both sides, it becomes a^2 = b^2 + n(2bc + nc^2) so, a^2 = b^2 (mod n) Q. E. D
is it ok?
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April 23rd, 2012, 05:13 AM   #5
 
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Re: how to prove that if a=b(mod n) then ....

hey thanks man, i guess my maths suck cause i dont quite get it, but i'll try get my head round it
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April 23rd, 2012, 05:41 AM   #6
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Re: how to prove that if a=b(mod n) then ....

Ok, then let me try again, but this time i will try to be more descriptive,
a = b (mod n) means a - b is divisible by n
then a - b = n*c where c is an nonzero integer.
then a = b + n*c
Now squaring both sides , it becomes
a^2 = (b + n*c)^2
By the squaring formula,
a^2 = b^2 + 2b*n*c + (n^2)*(c^2) = b^2 + n*(2b*c + (c^2)*n)
We know that a,b,c,n are integers. So, 2b*c + (c^2)*n is an integer. It is also nonzero because c?0.
then 2b*c + (c^2)*n = p where p is a non zero integer.
So, a^2 = b^2 + n*p = b^2 (mod n)
Q. E. D
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April 23rd, 2012, 05:52 AM   #7
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Re: how to prove that if a=b(mod n) then ....

a == b
a - b == 0
(a - b)(a + b) == 0(a + b) == 0
(a^2 - b^2) == 0
a^2 == b^2
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April 23rd, 2012, 06:01 AM   #8
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Re: how to prove that if a=b(mod n) then ....

Quote:
Originally Posted by The Chaz
a == b
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April 23rd, 2012, 06:16 AM   #9
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Re: how to prove that if a=b(mod n) then ....

Quote:
Originally Posted by mathbalarka
Quote:
Originally Posted by The Chaz
a == b
== is commonly used to mean when you don't have access to special characters.
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April 23rd, 2012, 06:23 AM   #10
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Re: how to prove that if a=b(mod n) then ....

Didnt got it either, am i missing something? If a == b then how it is related to the original question?
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