# Addition or subtraction in GCD and LCM

Suppose that we have two integers $a$ and $b$. Now say that $G = \gcd(a,b)$ and $L = \mathrm{lcm}(a,b)$. Now the value of $G$ and $L$ is given and another integer $c$'s value is given. How can we find $\gcd(a+c,b+c)$ and $\mathrm{lcm}(a+c,b+c)$ from $G$, $L$ and $c$?

What if we have $n$ arbitrary numbers. I know the GCD and LCM of those numbers but not the actual values of those numbers. Now I want to add $c$ with all of those numbers, what will be the new GCD, LCM of those numbers?

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Do your variables $c$ and $C$ refer to the same number? In that case please consistently use the same letter. – joriki Oct 22 '12 at 19:42
The gcd(a+c,b+c) is a divisor of $a-b$. Therefore you have a finite number of possibilities as $c$ varies. – PAD Oct 22 '12 at 20:00
I believe there is a theorem (a special case of this) about Mersenne numbers which states that $\gcd ({2^n} - 1,{2^m} - 1) = {2^{\gcd (n,m)}} - 1$. – glebovg Oct 22 '12 at 20:25

@PAD has indicated that $\gcd(a +c , b + c)$ is a divisor of $a-b$. Let us show that all divisors of $a - b$ occur as $\gcd(a +c , b + c)$, for a suitable choice of $c$.
Let $\lambda \mid a - b$. Choose $c = \lambda - b$, so that $b + c = \lambda$, and $a + c = (b + c) + (a - b)$ is divisible by $\lambda$. It follows that $\gcd(a +c , b + c) = \lambda$.
Of course $\gcd (ka,kb) = |k|\gcd (a,b)$ for some integer $k \ne 0$ is easy to prove, but I do not think is a generalization you seek because this would mean the the gcd of any two numbers is somehow generated by $\gcd (a,b)$, where $a$ and $b$ are arbitrary. That does not seem likely. Unless $a$ and $b$ are coprime, which is trivial. The same goes for lcm.