# How to prove that $gcd(a+b,a-b) \ge gcd(a,b)$?

How to prove that $gcd(a+b,a-b) \ge gcd(a,b)$?
I am confused, is this related to Bezout's theorem?

• Just asking shouldn't it be: $gcd(a+b, a-b) \ge gcd(a,b)$? Tried using $a=24$, $b=2$. Mar 25 '17 at 9:39
• Yes, thanx, if u want u can edit Mar 25 '17 at 9:43

If $d$ divides $a+b,a-b$ both

$d$ must divide $a+b+(a-b)=2a$ and $a+b-(a-b)=2b$

$\implies d$ must divide $(2a,2b)=2(a,b)$

Conversely, if $n=(a,b); n$ must divide $a\pm b\implies n$ must divide $(a+b,a-b)$

• There should be a correction in the second line your of answer sir. Mar 25 '17 at 9:43
• @MayankM., Thanks Mar 25 '17 at 9:44
• I did not get it. Mar 25 '17 at 9:47
• @AmitayasBanerjee, Trying checking for (odd, odd);(odd, even);(even, even) values of $a,b$ Mar 25 '17 at 9:51
• Thanx, I got it Mar 25 '17 at 9:58

Although the question has been solved I just wanted to try this approach. I'd like to read some comment about it.
Let $$a=a_1^{\alpha_1} \cdot\space\dots\space\cdot a_n^{\alpha_n}$$
$$b = b_1^{\beta_1} \cdot\space\dots\space\cdot b_m^{\beta_m}$$

Now, let $HCF (a,b) = \lambda$.
$\implies a-b = \lambda [(a_i^{x_1} \cdot\space\dots\space\cdot a_{i+k}^{x_k}) - (b_j^{y_1} \cdot\space\dots\space\cdot b_{j+k}^{y_k})]$
$\implies a+b = \lambda [(a_i^{x_1} \cdot\space\dots\space\cdot a_{i+k}^{x_k}) + (b_j^{y_1} \cdot\space\dots\space\cdot b_{j+k}^{y_k})]$

Let, $p=[(a_i^{x_1} \cdot\space\dots\space\cdot a_{i+k}^{x_k}) - (b_j^{y_1} \cdot\space\dots\space\cdot b_{j+k}^{y_k})]$,
$q=[(a_i^{x_1} \cdot\space\dots\space\cdot a_{i+k}^{x_k}) + (b_j^{y_1} \cdot\space\dots\space\cdot b_{j+k}^{y_k})]$.
$\implies HCF(a+b, a-b) = \lambda \cdot HCF(p,q) = HCF(a,b)\cdot HCF(p,q)$

Therefore, $HCF(a+b, a-b) \ge HCF(a,b)$.

• This is really what I was looking for! Mar 26 '17 at 14:51