Find the integer, \(n\), that satisfies \(n^2<33127<(n+1)^2\). Find also a small integer \(m\) such that \((n+m)^2-33127\) is a perfect square. Hence express \(33127\) in the form \(pq\), where \(p\) and \(q\) are integers greater than \(1\).

Since \(182^2=33124\) and \(183^2=33489\), we have \(n=182\).

We have \(184^2-33127=729=27^2\), so \(m=2\) is a small integer that works.

Notice that we are close to the difference of two squares here, which could be helpful for factorising.

Therefore \(184^2-27^2=33127\), so \(33127=(184-27)\times(184+27)\), which implies that \(33127=157\times211\).

By considering the possible factorisations of \(33127\), show that there are exactly two positive values of \(m\) for which \((n+m)^2-33127\) is a perfect square, and find the other value.

Since \(157\) and \(211\) are both prime numbers, the only other factorisation of \(33127\) is \(1\times33127\).

If \((182+m)^2 - 33127 = k^2\), where \(k\) and \(m\) are positive integers, then \(33127\) is the difference of two squares, so \[33127 = (182+m)^2 - k^2 = (182+m-k)(182+m+k).\]

So \(182+m-k\) and \(182+m+k\) must be factors of \(33127\).

Thus we either have \(182+m+k = 211\) and \(182+m-k = 157\) (which we have explored already) , or \(182+m+k = 33127\) and \(182+m-k=1.\)

The first pair of simultaneous equations solve to give \(m = 2, k = 27\), as we have seen.

The second pair give \(m = 16382, k = 16563\).

(Considering the case when \(k\) is negative gives us no new solutions).