(a) Find a solution to the diophantine equation $2X^2+2Y^2=Z^2$; hence find a solution for rational numbers of the form $2x^2+2y^2=1$.

We have $2X^2+2Y^2 \equiv Z^2\pmod{2} \implies (X,Y,Z) =c(1,1,2), ~c \in\mathbb{Z}$ is a solution. And $(1/2, 1/2, 1)$ is a possible solution to the rational equation.

(b) Find all rational solutions of the equation $2x^2+2y^2=1$; hence find all integer solutions to $2X^2+2Y^2=Z^2$

I'm not sure about this.

I know we can solve $y = m(x-1/2)+\frac{1}{2}$ and $2x^2+2y^2=1$ to get

$\displaystyle x= \frac{m^2-2m-1}{2(m^2+1)}, y= \frac{1-2m-m^2}{2(m^2+1)}$

(1) I don't know if $m$ is rational. So I'm not sure if I've found any rational solutions.

(2) If I force $\displaystyle m=\frac{p}{q}$ be rational I get

$\displaystyle x = \frac{(p^2-2pq-q^2)}{ 2(p^2 + q^2)} $, $\displaystyle y = \frac{-(p^2 + 2 p q - q^2)}{ 2(p^2 + q^2)}$.

Then solution to the integer equation $2X^2+2Y^2=Z^2$ is:

$\left(x, y, z \right) = \left(p^2-2pq-q^2, -(p^2 + 2 p q - q^2), 2(p^2 + q^2) \right)$.

Does this sound right? Can I just assume that $m$ is rational? Why can one rational solution generate all rational solutions (if it does)? Could I have gotten the general solution for the integer version at the beginning with just modular arithmetic?


Think about what you are doing here. $2x^2+2y^2=1$ is a circle, and you have found a rational point $P$ on it. Two observations link the algebra and the geometry:

A straight line between two rational points in the plane has a rational slope ($m$ must be rational).

If you have a straight line through $P$ with rational slope $m$ it will cut the circle at another rational point (or be a tangent). This is a consequence of Vieta's formulae - the quadratic for the circle has rational coefficients, so the sum of roots of the equation for $x$ (or $y$) is rational. One root is rational so both are.

So for arbitrary $m$ you parametrise all the lines through $P$. Constraining to rational $m$ gives you all the lines which cut the circle at a second rational point (apart from one corresponding to $m=\infty$; what happens with the tangent at $P$?).

You are allowed to constrain $m$ to get the rational points. I hope this helps you to see why it works.

  • $\begingroup$ Thank you. I now understand the point of this exercise a lot better I think. I wonder whether I could do something like this: set each coordinate of my parametrisation in terms of $m$ -- that's $[(m^2-2m-1)/[2(m^2+1)],-(m^2-2m-1)/[2(m^2+1)]) $ -- to zero. It gives $m=1 \pm \sqrt{2}$ this takes $(1 \pm \sqrt{2}, 1 \pm \sqrt{2}) \mapsto (0,0)$ which is not on the circle; thus $1 \pm \sqrt{2}$ is irrational; for if it were rational it would be taken to a rational point on the curve? $\endgroup$ – Notsredt Jul 15 '19 at 9:07

You have found one rational solution $(x,y)=(\frac12,\frac12)$ to $$ 2|x+iy|^2=1 $$ This you can now use to transform any other solution as in $$ 2=|1-i|^2\implies |(1-i)(x+iy)|^2=1, $$ that is, $(x+y, y-x)$ is a rational point on the unit circle. Now these are all parametrized by Pythagorean triples $a+ib=(p+iq)^2$, $c=|p+iq|^2$. All together, $$ x=\frac{p^2-q^2-2pq}{2(p^2+q^2)},~~~y=\frac{p^2-q^2+2pq}{2(p^2+q^2)}. $$ All reflections and rotations can be obtained via permutation and sign-flips in the pair $(p,q)\in\Bbb Z^2$.


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