Problem
NT-B1-M02-P010 Sum and Difference
#10
★★☆☆☆ Level 2 of 5
Let \(\gcd(a,b)=1\). Prove that \(\gcd(a+b,a-b)\) divides \(2\).
A common divisor of the sum and difference divides their sum and difference.
Let \(d\mid a+b\) and \(d\mid a-b\). Then \(d\mid (a+b)+(a-b)=2a\) and \(d\mid(a+b)-(a-b)=2b\). Since \(\gcd(a,b)=1\), no odd prime can divide \(d\). Therefore \(d\mid2\).
A very common template for sums and differences.