# GCD for three numbers (number theory)

For odd natural numbers $$a,b,c$$, prove that: $$\gcd \left( \frac{a+b}{2}, \frac{b+c}{2}, \frac{a+c}{2} \right) = \gcd(a, b, c).$$

How can we deal with the fact that: $$a = \frac{a + c}{2} + \frac{a + b}{2} - \frac{b + c}{2}.$$

Well, we know that if b > a, then gcd(a, b) = gcd(a, b-a). Using this property I resulted $$\gcd \left( \frac{a+b}{2}, \frac{b+c}{2}, \frac{a+c}{2} \right)$$ in the form $$\gcd \left( b, \frac{b+c}{2}, \frac{a-b}{2} \right)$$ It seems to me that all that remains is to achieve a and c from the $$\frac{b+c}{2}, \frac{a-b}{2}$$ by performing arithmetic operations. All my attempts to do this were unsuccessful...

It seems to me that a slightly different approach is needed here. Therefore, if you have any ideas or suggestions for a solution, please write them, I will be grateful for any hint!

• Nice question, but you need to show your attempts to answer it... Jan 19 at 15:07
• please put your attempts in the body of the question, not in the comments. Thanks. Jan 19 at 15:29
• $\frac{a+b}{2}-\frac{b+c}{2}+\frac{a+c}{2}=a$ and similar operations allow you to get $b$ and $c$. Therefore, any divisor of the three fractions on the left divides each of the values on the right. The other direction is easier. That's all you need. Jan 19 at 16:40

The fact that you can express $$a$$, and analogously $$b$$ and $$c$$, as a sum and difference of $$r=\frac{a+b}2$$, $$s=\frac{b+c}2$$ and $$t=\frac{a+c}2$$ shows that any common divisor of $$r$$, $$s$$ and $$t$$ is also a common divisor of $$a$$, $$b$$ and $$c$$. For the other direction, since $$a$$, $$b$$ and $$c$$ are odd, their common divisor isn’t divisble by $$2$$, so e.g. $$r=\frac{a+b}2$$ is divisible by all common divisors of $$a$$ and $$b$$ since the $$2$$ in the denominator doesn’t divide any of them.