prove that for all integers a and c, if a is odd, c0, c devides a and c devides a+2, then c= 1.

Question

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.

zzr0ck3r · Answer

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

zarkon · Answer

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

zarkon · Answer

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

zzr0ck3r · Answer

ahh the second one is more along the lines of what we are doing in class.

zzr0ck3r · Answer

nice work:)