Problem:
Let q=23p−5​ where p is an odd prime, and let
Sq​=2⋅3⋅41​+5⋅6⋅71​+…+q(q+1)(q+2)1​
Prove that if p1​−2Sq​=nm​ for integers m and n, then m−n is divisible by p.
Solution:
Solution by Titu Andreescu: We have
k(k+1)(k+2)2​​=k(k+1)(k+2)(k+2)−k​=k(k+1)1​−(k+1)(k+2)1​=k1​−k+11​−(k+11​−k+21​)=k1​+k+11​+k+21​−k+13​​
Hence
2Sq​​=(21​+31​+41​+…+q1​+q+11​+q+21​)−3(31​+61​+…+q+11​)=(21​+31​+…+23p−1​1​)−(1+21​+…+2p−1​1​)​
and so
1−nm​​=1+2Sq​−p1​=2p+1​1​+…+p−11​+p+11​+…+23p−1​1​=(2p+1​1​+23p−1​1​)+…+(p−11​+p+11​)=(2p+1​)(23p−1​)p​+…+(p−1)(p+1)p​​
Because all denominators are relatively prime with p, it follows that n−m is divisible by p and we are done.
This problem was suggested by Titu Andreescu.
The problems on this page are the property of the MAA's American Mathematics Competitions