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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.

Solution:

Triangular Numbers Terms Number of terms Formula
3 1+2 2 2/2 (3)
6 1+2+3 3 3/2 (4)
10 1+2+3+4 4 4/2 (5)
15 1+2+3+4+5 5 5/2 (6)
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.