Biren Patel, Heathland School, Hounslow, England; Andrei, School no. 205, Bucharest, Romania and Ang Zhi and Chai from River Valley High School, Singapore sent in very good solutions. Here is Chai solution:
Prove that if n is a triangular number then 8n+1 is a square number.
Prove, conversely, that if 8n+1 is a square number then n is a triangular number.
|Triangular Numbers||Terms||Number of terms||Formula|
|n||1+2+3+4+5+... K||K||K/2 (K+1)|
Substitute n = K/2 (K+1) into 8n + 1
|8n + 1||=||8 [K/2 (K+1)] + 1|
|=||8/2 (K 2 + K) + 1|
|=||4K 2 + 4K + 1|
|=||(2K + 1) 2|
Therefore, if n is a triangular number then 8n+1 is a square number.
To prove the converse, let X 2 = 8n + 1 be a square number.
As 8 is an even number, 8n will always be an even number. If 8n is an even number, then 8n + 1 will always be an odd number. X cannot be even because the square of an even number is even.
Hence X= (2k + 1) represents an odd number where k can be any whole number.
(2k + 1) 2 = 8n + 1
4k 2 + 4k+ 1 = 8n + 1
4k 2 + 4k + 1 - 1 = 8n
4/8(k 2 + k) = n
(k 2 + k)/2 = n
Therefore, n = k/2 (k+1) so n is a triangular number.