1 problem found
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 \(16564^2 - 33\,127 = 16563^2\)