prove that for all integers a and c, if a is odd, c>0, c devides a and c devides a+2, then c= 1.
Hey man I see you typing and I have to go to a party but thank you so much and I will look when I get back in a few hours. ty
if \(a\) is odd then \(a=2k-1\) for some integrer k then \(a+2=2k+1\) notice that \[(k)\times(2k-1)+(-(k-1))\times(2k+1)=1\] thus the GCD(\(2k-1,2k+1\))=1 so the greatest common divisor is 1. since \(c\) divides both, \(c\) must be 1
I thought of a 2nd proof suppose \(c|a\) and \(c|a+2\) then \(a=k_1c\) and \(a+2=k_2c\) where \(k_1,k_2\) are both odd (since \(a,a+2\) are odd) clearly \(k_2>k_1\) and their minimum difference is 2 thus \(k_2-k_1>1\) then \(a=k_1c\Rightarrow a+2=k_1c+2\) so \(k_1c+2=k_2c\) then \(2=k_2c-k_1c=(k_2-k_1)c\) since \(k_2-k_1>1\) and \(k_2-k_1,c\) are integers we must have that c=1
ahh the second one is more along the lines of what we are doing in class.
nice work:)
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