We have to find the integers $m$ and $n$ which will satisfy the given condition: $$m^2-n^2=1111.$$ What could be the answer and how? i tried using trial and error and that took a long time.
N.B. This is an answer to a slightly different question, hopefully for instructional purposes. It gives the necessary ideas, but not the solution.
Let's do it for a different number. Replace $1111$ by $11$. Then
$$11 = m^2 - n^2 = (m - n)(m + n)$$
Now the factors of $11$ are $1$ and $11$, so we have
$$m - n = 1 \quad\quad m + n = 11$$ or $$m - n = 11 \quad\quad m + n = 1$$
or the cases where $1\cdot 11$ is replaced by $(-1) \cdot (-11)$. For example, the first case leads to a solution $m = 6, n = 5$, and the difference of squares representation $$6^2 - 5^2 = 11$$
Now $1111$ is not prime, so it's more complicated. But this should at least get you started.