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2014 Paper 1 Q1
D: 1500.0 B: 1500.0
proof number theory difference of two squares parity factorisation odd and even prime numbers counting arguments

All numbers referred to in this question are non-negative integers.

  1. Express each of the numbers 3, 5, 8, 12 and 16 as the difference of two non-zero squares.
  2. Prove that any odd number can be written as the difference of two squares.
  3. Prove that all numbers of the form \(4k\), where \(k\) is a non-negative integer, can be written as the difference of two squares.
  4. Prove that no number of the form \(4k+2\), where \(k\) is a non-negative integer, can be written as the difference of two squares.
  5. Prove that any number of the form \(pq\), where \(p\) and \(q\) are prime numbers greater than 2, can be written as the difference of two squares in exactly two distinct ways. Does this result hold if \(p\) is a prime greater than 2 and \(q=2\)?
  6. Determine the number of distinct ways in which 675 can be written as the difference of two squares.

Solution:

  1. \(\,\) \begin{align*} && 3 &= 2^2 - 1^2 \\ && 5 &= 3^2 - 2^2 \\ && 8 &= 3^2 - 1^2 \\ && 16 &= 5^2 - 3^2 \end{align*}
  2. Suppose \(n = 2k+1\), then \(n = (k+1)^2 - k^2\)
  3. Suppose \(n = 4k\) then \(n = (2k+1)^2 - (2k-1)^2\)
  4. All squares leave a remainder of \(0\) or \(1\) on division by \(4\). Therefore the difference can leave a remainder of \(0\), \(1\), \(-1 \equiv 3\), none of which are \(2\).
  5. Suppose \(n = pq = a^2 - b^2\) with \(a > b\) ie \((a-b)(a+b) = pq\). Since \(p\) is prime, \(p \mid (a-b)\) or \(p \mid (a+b)\). Similarly for \(q\). Suppose also (wlog) that \(p > q\) Since the factors of \(pq\) are \(1, p, q, pq\) then \(a-b = 1, p\) (which are two possibilities) and \(a+b = pq, q\), ie \(a = \frac{1+pq}{2}, \frac{p+q}{2}\) and \(b = \frac{pq-1}{2}, \frac{p-q}{2}\) \begin{align*} && pq &= \left ( \frac{1+pq}{2} \right)^2- \left ( \frac{1-pq}{2} \right)^2 \\ &&&= \left ( \frac{p+q}{2} \right)^2- \left ( \frac{p-q}{2} \right)^2 \\ \end{align*} Where everything is an integer since \(p\) and \(q\) are odd. If we have \(p > 2\) and \(q = 2\) then \(p\) is odd and the number has the form \(4k+2\) which cannot be expressed as the difference of two squares.
  6. \(675 = 3^3 \cdot 5^2\), each factor pair of \(675\) will lead to a different solution of \(675 = a^2-b^2\), since we will have an equation \(a-b = X, a+b = Y\) where \(X, Y\) are both odd. Therefore there are as many solution as (half) the number of factors, ie \(4 \times 3 = 12\)

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2006 Paper 1 Q1
D: 1500.0 B: 1516.0
number theory perfect squares factorisation difference of two squares integer arithmetic systematic search

Find the integer, \(n\), that satisfies \(n^2 < 33\,127< (n+1)^2\). Find also a small integer \(m\) such that \((n+m)^2-33\,127\) is a perfect square. Hence express \(33\,127\) in the form \(pq\), where \(p\) and \(q\) are integers greater than \(1\). By considering the possible factorisations of \(33\, 127\), show that there are exactly two values of \(m\) for which \((n+m)^2 -33\,127\) is a perfect square, and find the other value.

Solution: \begin{align*} 180^2 &= 32400 \\ 181^2 &= 32761 \\ 182^2 &= 33124 \\ 183^2 &= 33489 \\ 184^2 &= 33856 \end{align*} Therefore \(182^2 < 33\,127 < (182+1)^2\). and \((182+2)^2 - 33\,127 = 729 = 27^2\). Therefore \(33\,127 = 184^2 - 27^2 = 211 \times 157\). (Note both of these numbers are prime). Suppose \((n+m)^2 - 33\,127 = k^2\) then \(33\,127 = (n+m)^2-k^2 = (n+m-k)(n+m+k)\). Since there are only two factorisations of \(33\,127\) into positive integer factors with one factor larger than the other, the other factorisation must be: \(n+m+k = 33\,127, n+m-k = 1 \Rightarrow k = \frac{33\, 126}{2} = 16563\), ie \(16564^2 - 33\,127 = 16563^2\)

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