MHB Solve Diophantine Equation: $m^2 - n^2 = 105$

[Jkawa]

Find all solutions to $m^2 - n^2 = 105$, for which both $m$ and $n$ are integers.

 

[Prove It]

Re: Help needed for problem

Find all solutions to $m^2 - n^2 = 105$, for which both $m$ and $n$ are integers.

Notice $\displaystyle \begin{align*} m^2 - n^2 = \left( m - n \right) \left( m + n \right) \end{align*}$. So what are the factors of 105 that could be possible candidates for m-n and m+n?

 

[Jkawa]

Re: Help needed for problem

Notice $\displaystyle \begin{align*} m^2 - n^2 = \left( m - n \right) \left( m + n \right) \end{align*}$. So what are the factors of 105 that could be possible candidates for m-n and m+n?

Factors of 105 are 1, 3, 5, 7, 15, 21, 35, 105. I'm sort of confused on the use of these though. I don't see a combination that could satisfy the equation.

 

[MarkFL]

Let's start with the pair $1\cdot105$.

First, compute the mean:

$$\frac{1+105}{2}=53$$

Now, find the difference between the mean and the smaller factor:

$$53-1=52$$

Thus:

$$(53+52)(53-52)=105$$

Proceed in like manner for the remaining pairs. :)

 

[Jkawa]

Upon inspection, I only see that the only solution to this is $m = 53$ and $n = 52$ because the rest of the factors do not equal 105. Am I correct?

 

[MarkFL]

Upon inspection, I only see that the only solution to this is $m = 53$ and $n = 52$ because the rest of the factors do not equal 105. Am I correct?

No, you can do the same with $3\cdot35$.

$$\frac{3+35}{2}=19$$

$$19-3=16$$

$$(19+16)(19-16)=105$$

Try the other pairs. :D

 

[Jkawa]

No, you can do the same with $3\cdot35$.

$$\frac{3+35}{2}=19$$

$$19-3=16$$

$$(19+16)(19-16)=105$$

Try the other pairs. :D

Oh my calculation was completely off, I was following the exact same algorithm as you posted lol. Thanks!

 

[MarkFL]

Don't neglect the negative pairs of integer factors, for example:

$$(-1)(-105)$$

$$\frac{(-1)+(-105)}{2}=-53$$

$$-53-(-1)=-52$$

$$(-53+(-52))(-53-(-52))=105$$