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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1$\begingroup$ what are the factors of 1111?? $\endgroup$– JasserDec 22, 2015 at 8:26
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1$\begingroup$ got it ! excellent $\endgroup$– Aditya BidwaiDec 22, 2015 at 8:32
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$\begingroup$ $56^2-45^2=(56+45)(56-45)=101\times 11=1111.$ $\endgroup$– user249332Dec 22, 2015 at 8:34
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$\begingroup$ Similar type of problem: GRE Quantitative Problem: integers such that $P^2 - Q^2 = 1155$ $\endgroup$– Martin SleziakOct 27, 2019 at 6:03
2 Answers
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
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1$\begingroup$ This does not answer the question as you formulated it. You are missing negative solutions, i.e. $3$ other solutions. $\endgroup$– user228113Dec 22, 2015 at 8:40
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1$\begingroup$ in negative case the answer would be flipped $\endgroup$ Dec 22, 2015 at 8:41
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3$\begingroup$ @AdityaBidwai, There are four solutions in terms of (m,n) you have found: (m,n), (m,-n), (-m,n), (-m,-n). $\endgroup$– Galc127Dec 22, 2015 at 8:45
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1$\begingroup$ Actually, there are more solutions since $556^2-555^2 = (556+555)(556-555) = 1111 \cdot 1 = 1111$. $\endgroup$ Dec 22, 2015 at 9:29