# Twice the square of the first subtracted from the square of the second is -167, what are the two integers?

Jul 1, 2015

Even if we assume the integers are both positive, there are an infinite number of solutions to this question. The minimal (positive) values are
$\left(11 , 12\right)$

#### Explanation:

If the first integer is $x$ and the second integer is $y$

${y}^{2} - 2 {x}^{2} = - 167$

${y}^{2} = 2 {x}^{2} - 167$

$y = \pm \sqrt{2 {x}^{2} - 167}$
$\textcolor{w h i t e}{\text{XXXX}}$(from here on, I will limit my answer to positive values)

if $y$ is an integer
$\Rightarrow 2 {x}^{2} - 167 = {k}^{2}$ for some integer $k$

We could limit our search by noting that $k$ must be odd.

Since $x$ is an integer
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{{k}^{2} - 167}{2}$ must also be an integer

Unfortunately there are lots of solutions for $k$ that satisfy the stated conditions:

{:(k,,first,second), (11,,12,11), (15,,14,15), (81,,58,81), (101,,72,101), (475,,336,475), (591,,418,591) :}
are the values that I found for $k < 1000$
and all of these satisfy the given conditions.

(...and, yes, I know $k = y$).