The Computer Laboratory

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.

The above links are provided as is. They are not affiliated with the Climb Foundation unless otherwise specified.


  • 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.


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 \(n = \frac{2k(2k+2)}{8} = \frac{k(k+1)}{2} = \sum_{i=0}^{k}{i}.\)

If you have queries or suggestions about the content on this page or the CSAT Practice Platform then you can write to us at oi.footasc@sulp.ecitcarp. Please do not write to this address regarding general admissions or course queries.