# Practice Paper 3 Question 6

A positive integer $$n$$ is said to be triangular if $$n =\sum_{i=0}^{k}{i}$$ for some positive integer $$k.$$ Given $$8n+1$$ is a square number, show that $$n$$ is triangular.

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## Hints

• Hint 1
Solve $$\sum_{i=0}^{k}{i}.$$
• Hint 2
The condition can be written as $$8n+1 = m^2,$$ for some natural $$m$$.
• Hint 3
Manipulate the above to a form that allows factorization.\$
• Hint 4
If $$(m-1)(m+1)$$ is divisible by $$8,$$ what values can $$m$$ take?
• Hint 5
Would $$(m-1)(m+1)$$ be divisible by $$8$$ if $$m$$ was even?
• Hint 6
Use a substitution for $$m$$ to express that $$m$$ is odd.

## Solution

A positive integer $$n$$ is said to be triangular if $$n =\sum_{i=0}^{k}{i}=\frac{k(k+1)}{2}.$$ Since $$8n+1$$ is a square number, there exists a positive integer $$m$$ such that $$8n + 1 = m^2.$$ Using the difference between two squares, we can factorize to get $$8n = m^2-1 = (m-1)(m+1),$$ or that $$n=\frac{(m-1)(m+1)}{8},$$ which means that $$(m-1)(m+1)$$ must be divisible by 8. This is true if and only if $$m$$ is odd, since for every consecutive even numbers, one is a multiple of $$4.$$ Substituting $$m = 2k+1$$ gives us $$n = \frac{2k(2k+2)}{8} = \frac{k(k+1)}{2} = \sum_{i=0}^{k}{i}.$$

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