Problem source

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 source

\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 $16564^2 - 33\,127 =  16563^2$