All solutions to equation problem

[tmt1]

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?

 

[Prove It]

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 $\displaystyle \begin{align*} m + n = 8 \end{align*}$?