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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!

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  • $\begingroup$ Nice question, but you need to show your attempts to answer it... $\endgroup$ Jan 19 at 15:07
  • $\begingroup$ please put your attempts in the body of the question, not in the comments. Thanks. $\endgroup$ Jan 19 at 15:29
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    $\begingroup$ $\frac{a+b}{2}-\frac{b+c}{2}+\frac{a+c}{2}=a$ 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. $\endgroup$
    – John Douma
    Jan 19 at 16:40

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John Douma has basically answered the question in a comment.

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.

Since all common divisors on one side are also common divisors on the other side, the greatest common divisors coincide.

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  • $\begingroup$ Thanks for the answer! But I solved the problem, without using John Douma's comment, and the teacher said that the solution was correct. $\endgroup$ Jan 28 at 18:24
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    $\begingroup$ @JacobsMonarch: In that case please post your solution as an answer for the benefit of future readers. $\endgroup$
    – joriki
    Jan 28 at 22:38

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