MHB All solutions to equation problem

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The equation \( m^2 - n^2 = 56 \) can be factored using the difference of squares, leading to \( (m - n)(m + n) = 56 \). Several factor pairs of 56 are identified, including \( (1, 56), (2, 28), (4, 14), (7, 8) \). Each pair can be set as \( m - n \) and \( m + n \) to find positive integer solutions for \( m \) and \( n \). The possible combinations yield valid solutions, confirming that the equation can be solved through systematic pairing of factors. The discussion emphasizes the importance of recognizing all factor pairs to derive the complete set of solutions.
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Find all solutions to the equation $m^2 - n^2 = 56$ for which $m$ and $n$ are positive integers.

So, I find that $56 = 7\cdot8 = 28\cdot2 = 56\cdot1$.

Then, I do not know how to proceed from here.

Any suggestions?
 
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tmt said:
Find all solutions to the equation $m^2 - n^2 = 56$ for which $m$ and $n$ are positive integers.

So, I find that $56 = 7\cdot8 = 28\cdot2 = 56\cdot1$.

Then, I do not know how to proceed from here.

Any suggestions?

I would use the difference of two squares to help out here:

$\displaystyle \begin{align*} m^2 - n^2 &= 56 \\ \left( m - n \right) \left( m + n \right) &= 56 \end{align*}$

and as you have established, the possible combinations are $\displaystyle \begin{align*} 56 \cdot 1 , \, 28 \cdot 2 , \, 7 \cdot 8 \end{align*}$ and also $\displaystyle \begin{align*} 14 \cdot 4 \end{align*}$ which you missed.

So is it possible to find two numbers so that $\displaystyle \begin{align*} m - n = 1 \end{align*}$ and $\displaystyle \begin{align*} m + n = 56 \end{align*}$?

How about where $\displaystyle \begin{align*} m - n = 2 \end{align*}$ and $\displaystyle \begin{align*} m + n = 28 \end{align*}$?

How about where $\displaystyle \begin{align*} m - n = 4 \end{align*}$ and $\displaystyle \begin{align*} m + n = 14 \end{align*}$?

How about where $\displaystyle \begin{align*} m - n = 7 \end{align*}$ and $\displaystyle \begin{align*} m + n = 8 \end{align*}$?
 
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