finding solutions by factoring

Question

How would you find the integer solutions to $a^2-b^2=16$?

I know that the factors of $16$ are $8*2,$ $4*4,$ and $16*1.$ How would I use this?

I know that $a^2-b^2=(a+b)(a-b)=16,$ but how would you find to the solutions to it? I know that you can plug in numbers to find the answers, but is there another way of doing this?

Answer 1

We wish to find integer solutions for $a$ and $b$ such that $a^2-b^2=16$.

Factoring the expression we obtain $(a+b)(a-b)=16$.

This means that $(a+b)|16$ and $(a-b)|16$.

From this we can find a set $(a+b)\in \{-16,-8,-4,-2,-1,1,2,4,8,16\}$.

So we obtain the equations

$a+b=-16 \implies a-b=-1$

$a+b=-8 \implies a-b=-2$

$a+b=-4 \implies a-b =-4$

$a+b =-2 \implies a-b =-8$

$a+b = -1 \implies a-b=-16$

$a+b= 1 \implies a-b=16$

$a+b=2 \implies a-b = 8$

$a+b=4 \implies a-b = 4$

$a+b=8 \implies a-b = 2$

$a+b=16 \implies a-b=1$

Now for each of these there is a corresponding matrix (e.g. the first set of equations has been done below: \begin{align*} \left(\begin{array}{cc|c} 1 & 1 &-16\\ 1 & -1 &-1\\ \end{array}\right)\end{align*} Which we may factor into row echelon form \begin{align*} \left(\begin{array}{cc|c} 1 & 1 &-16\\ 0 & -2 & 15\\ \end{array}\right)\end{align*}

To get the solution of $b=-7.5$ and $a=-8.5$.

This is not an integer solution so we may discard it. Now continue this process for the remaining 9 pairs of equations.

Answer 2

Introduce a temporary pair of new variables: Call $u=a+b$ and $v=a-b$. Notice that if $a$ and $b$ are integers, then $u$ and $v$ will be too. Notice also that if you know values for $u$ and $v$ it is a straightforward matter to figure out what $a$ and $b$ are.

Now the equation to solve is $uv=16$. So just list all of the pairs of integers that multiply to be $16$: $1 \cdot 16$, $2 \cdot 8$, etc. Those give you the integer values of $u$ and $v$; use those to find the values of $a$ and $b$.