# A Curious Investigation

Standard

As I always say:

Mathematicians are those weird beasts who enjoy being surrounded by problems.

My current field of interest is Diophantine Equations (DE). Those who ever studied Number theory know about the classic Pythagorean Triplets,  equivalent to finding possible integer solutions of $x^2+y^2 = z^2$.

Also, it is a standard exercise (involving Method of Infinite Descent) in DE to prove  that$x^2+y^2 = 3 z^2$ has no integer solutions. But in this blog post I intend  to discuss following sibling of such degree two DE:

Solve $x^2+y^2 = 2z^2$ for integers.

Clearly, $z\neq 0$, I can divide whole equation by $z^2$ and denote, $X=x/z$ and $Y=y/z$ to get:

Solve $X^2+Y^2 = 2$ for rational numbers.

Observe that $(1,1)$ is a solution of given equation, then any other solution $(a,b)$ will lie with $(1,1)$ on a line with rational slope (or infinite slope, a vertical line, trivial case). Furthermore, every line through $(1,1)$ with rational slope will intersect with the quadratic curve in exactly two points. Because every quadratic equation has either no solutions or two real solutions, and we already know that $(1,1)$ is one solution.

To find all solutions, we first look at the vertical line case by substituting $X=1$  and seeing what two solutions you get.  One will be $Y=1$, and the other gives a solution (which is the same solution in this case). Next, we take a line with rational slope $m$ through $(1,1)$, so that (using slope-intercept form):

$Y=m(X-1)+1$

Now solve this line  and given curve $X^2+Y^2 = 2$ (which is circle of radius $\sqrt{2}$). We will get:

$X = \frac{m^2-2m -1}{m^2+1}$

$Y=\frac{-m^2-2m+1}{m^2+1}$

Where, $m \in \mathbb{Q}$, thus like $x^2+y^2=z^2$, $x^2+y^2 = 2z^2$ has infinite integer solution.

Ending note:

$x^2+y^2 = 2$ has only 4 integer solutions.

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