# Finding integers satisfying $m^2 - n^2 = 1111$

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.

• what are the factors of 1111?? Dec 22, 2015 at 8:26
• got it ! excellent Dec 22, 2015 at 8:32
• $56^2-45^2=(56+45)(56-45)=101\times 11=1111.$
– user249332
Dec 22, 2015 at 8:34
• Similar type of problem: GRE Quantitative Problem: integers such that $P^2 - Q^2 = 1155$ Oct 27, 2019 at 6:03

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.

$$(m+n)(m-n)=101\cdot11$$ therefore m=56 and n=45

• Now that was simple!! Dec 22, 2015 at 8:38
• This does not answer the question as you formulated it. You are missing negative solutions, i.e. $3$ other solutions.
– user228113
Dec 22, 2015 at 8:40
• in negative case the answer would be flipped Dec 22, 2015 at 8:41
• @AdityaBidwai, There are four solutions in terms of (m,n) you have found: (m,n), (m,-n), (-m,n), (-m,-n). Dec 22, 2015 at 8:45
• Actually, there are more solutions since $556^2-555^2 = (556+555)(556-555) = 1111 \cdot 1 = 1111$. Dec 22, 2015 at 9:29