prime numbers and greatest common divisor Question: For any prime numbers $p$ and $q$, one has $\text{gcd}(p+q, q) = \text{gcd}(p,q)$. Is this statement true?
Answer: True.
can someone explain why?
my reasoning: for any prime numbers, if an integer $q$ is added to it, the $\text{gcd}$ between $p+q$ and $q$ will still be $1$, since $q$ is still a prime number (prime = $1$(prime))
 A: $p$ and $q$ don't even have to be prime: $\gcd(a,b)=\gcd(a+b,b)$. The Euclidean algorithm is based on this property.
The reason for this  relies on this simple observation, easily checked: if $D(a,b)$ denotes the set of common divisors of $a$ and $b$, we have $D(a+b,b)=D(a,b)$, and more generally, for any integer $q$,
$$D(a+qb,b)=D(a,b).$$
This is quite obvious since any number which divides $a$ and $b$ divides $a+qb$ ($\supset$) and conversely, a number which divides $b$ and $a+qb$ divides $(a+qb)-qb=a$ ($\subset$).
A: We define the greatest common divisor (gcd) of two or more integers, when at least one of them is not zero, is the largest positive integer that divides the numbers without a remainder. 
More precisely, given $a, b \in \mathbb{Z}$, their gcd, let us denote it by $d$ is the positive integer such that 


*

*$d|a$ and $d|b$, (where notation $x|y$ means $x$ divides $y$)

*If $d^{'}$ is another positive integer which satisfies 1, that is $d^{'}$ also divides $a$ and $b$, then $d^{'}|d$


Now you want to show that $gcd(a,b)=gcd(a+b,b)$. 
So let us apply the definition. Let $\alpha= gcd(a,b)$, we will show that $gcd(a+b,b)=\alpha$. 
By definition $\alpha |a$ and $\alpha |b$, it implies that $\alpha |(a+b)$ so first condition is satisfied. You can check the second condition. Hence we have shown that $\alpha=gcd(a+b,b)$
