I believe this statement is true, but I've only tried one set of numbers (a = 8, b = 12). How would I go about proving this?
Asked
Active
Viewed 47 times
-1
-
3try two odd numbers. – lulu Sep 29 '15 at 01:22
-
1Used a = 35, b = 25. gcd(35,25) = 5, but gcd (60, 10) = 10. So looks like this statement is actually false. Thank you! – David Meredith Sep 29 '15 at 01:26
-
2You can, however, show that if $d=(a,b)$ then $(a+b,a-b)$ is either $d$ or $2d$. – lulu Sep 29 '15 at 01:30
-
1Simpler: $a=b=1$ – egreg Sep 29 '15 at 01:34
-
See also http://math.stackexchange.com/a/1445882/589. – lhf Sep 29 '15 at 01:55
1 Answers
0
Here is one way to prove lulu's comment, which is actually quite useful.
$\gcd(a+b,a-b) = \gcd((a+b)+(a-b),a+b) = \gcd(2a,a+b)$
${} \mid \gcd(2,a+b) \gcd(a,a+b) \mid 2 \gcd(a,b)$.
user21820
- 57,693
- 9
- 98
- 256