Question

If a and b are both odd positive integers with a > b, the prove that one of the integers $\dfrac{a+b}{2}$ and $\dfrac{a-b}{2}$ is even and one is odd.

Answer 1

Hint: Using Euclid's division lemma with b = 4, write all the possible forms of an integer n. Hence write the all possible configurations of a and b. Verify that in each case one of the integers $\dfrac{a+b}{2}$ or $\dfrac{a-b}{2}$ is an integer. Hence prove the result.

Complete step-by-step answer:
Let n be any integer.
From Euclid's division lemma, we have
n = 4q+r, $0\le r < 3$
Hence n is of one of the forms 4q, 4q+1, 4q+2 or 4q+3.
Since 4q and 4q+2 are even, we have any odd integer n is of one of the forms 4q+1 or 4q+3.
Case I: a = 4m+1 and b =4n+1.
We have $\dfrac{a+b}{2}=\dfrac{4m+1+4n+1}{2}=\dfrac{4(m+n)+2}{2}=2(m+n)+1$ which is odd
and $\dfrac{a-b}{2}=\dfrac{4m+1-4n-1}{2}=\dfrac{4\left( m-n \right)}{2}=2\left( m-n \right)$ which is even.
Hence one of the integers $\dfrac{a+b}{2}$ and $\dfrac{a-b}{2}$ is odd, and one is even.
Case II: a= 4m+1 and b = 4n+3.
We have $\dfrac{a+b}{2}=\dfrac{4m+1+4n+3}{2}=\dfrac{4m+4n+4}{2}=\dfrac{4\left( m+n+1 \right)}{2}=2\left( m+n+1 \right)$ which is an even integer.
and $\dfrac{a-b}{2}=\dfrac{4m+1-4n-3}{2}=\dfrac{2\left( 2m-2n-1 \right)}{2}=2\left( m-n \right)-1$ which is an odd integer.
Hence one of the integers $\dfrac{a+b}{2}$ and $\dfrac{a-b}{2}$ is odd, and one is even.
Case III: a = 4m+3 and b = 4n+1
We have $\dfrac{a+b}{2}=\dfrac{4m+3+4n+1}{2}=\dfrac{4\left( m+n+1 \right)}{2}=2\left( m+n+1 \right)$ which is even
and $\dfrac{a-b}{2}=\dfrac{4m+3-4n-1}{2}=2\left( m-n \right)+1$ which is odd
Hence one of the integers $\dfrac{a+b}{2}$ and $\dfrac{a-b}{2}$ is odd, and one is even.
Case IV: a=4m+3 and b = 4n+3
We have $\dfrac{a+b}{2}=\dfrac{4m+3+4n+3}{2}=2\left( m+n+1 \right)+1$ which is odd
and $\dfrac{a-b}{2}=\dfrac{4m+3-4n-3}{2}=2\left( m-n \right)$ which is even.
Hence one of the integers $\dfrac{a+b}{2}$ and $\dfrac{a-b}{2}$ is odd, and one is even.
Hence in all cases, one of the integers $\dfrac{a+b}{2}$ and $\dfrac{a-b}{2}$ is odd, and one is even.
Hence proved.

Note: Alternative Solution: Best method
We know that if $\dfrac{a-b}{2}$ and $\dfrac{a+b}{2}$ are either both even or both are odd, then $\dfrac{a-b}{2}+\dfrac{a+b}{2}$ is even.
Now since $\dfrac{a-b}{2}+\dfrac{a+b}{2}=\dfrac{2a}{2}=a$ which is odd, they cannot be both even or both odd.
Hence one of $\dfrac{a-b}{2}$ and $\dfrac{a+b}{2}$ is even, and one is odd.
Hence proved.