Problem
ALG-B3-M06-P010 Alternating Parity
#10
★★★☆☆ Level 3 of 5
Prove that there is no \(f:\mathbb Z\to\mathbb Z\) such that \(f(n+1)-f(n)=2n+1\) and all values \(f(n)\) have the same parity.
The difference of neighbouring values is odd.
For every \(n\), the number \(2n+1\) is odd. Hence \(f(n+1)\) and \(f(n)\) have opposite parity. Therefore all values of the function cannot have the same parity. Contradiction.
Short application of parity as an obstruction.