Review question

# Can we write 33127 as the difference of two squares? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6171

## Solution

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).