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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.

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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.

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

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  • $\begingroup$ Now that was simple!! $\endgroup$
    – Jasser
    Dec 22, 2015 at 8:38
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    $\begingroup$ This does not answer the question as you formulated it. You are missing negative solutions, i.e. $3$ other solutions. $\endgroup$
    – user228113
    Dec 22, 2015 at 8:40
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    $\begingroup$ in negative case the answer would be flipped $\endgroup$ Dec 22, 2015 at 8:41
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    $\begingroup$ @AdityaBidwai, There are four solutions in terms of (m,n) you have found: (m,n), (m,-n), (-m,n), (-m,-n). $\endgroup$
    – Galc127
    Dec 22, 2015 at 8:45
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    $\begingroup$ Actually, there are more solutions since $556^2-555^2 = (556+555)(556-555) = 1111 \cdot 1 = 1111$. $\endgroup$
    – JimmyK4542
    Dec 22, 2015 at 9:29

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