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.