# Find two positive rational solutions to a Diophantine problem

I need to answer this question using the Diophantine method. The question is: Find two numbers so that the square of either number, plus twice the other number, is also a square. Give two sets of positive, rational solutions.

I tried using the following equations:

x^2 + 2(2x+1) = (x+1)^2

(2x+1)^2 + 2x = (3x-1)^2

These didn't work, so I changed the (3x-1) a few times, and those didn't work. Am I missing something? Please help. Thanks.

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I interpret "Diophantine method" as the method of parameterizing conics (as well as some higher-degree curves that have points with sufficient multiplicity). See my answer below. Plug in rational values $s>0,t>0$ with $st<1$, to find suitable rational $x,y$. – ccorn Oct 30 '13 at 20:01
I know how to do what you did, but I need to use the ancient Greek method, which is what I can't figure out. I can only use one variable and the answer is almost always a pair of fractions. Thanks for the help, though. – Mike Perez Oct 30 '13 at 20:28

According to the problem's description, you are looking for $u,v,x,y\in\mathbb{Q}$, $0<x<u$, $0<y<v$ such that \begin{align} u^2-x^2 &= 2y \tag{1a} \\ v^2-y^2 &= 2x \tag{1b} \end{align} (Only the squares of $u$ and $v$ are needed, so we can restrict the signs of $u,v$ to be positive.)

Suppose for the moment that you have fixed a positive rational value for $y$. Then $\text{(1a)}$ describes a hyperbola in the $(u,x)$ plane. We will parameterize all points on that curve with positive rational coordinates. Of course, that parametrization will make $x$ (and $u$) depend not only on the new parameter, but on $y$ as well. By symmetry, from $\text{(1b)}$ we will find an analogous parametrization for $y$ (and $v$) also dependent on $x$. Together, the parametrizations can be taken as a system of equations for $(x,y)$ which we will solve.

For $\text{(1a)}$, Diophantus's method would be: Find one rational point $(u_0,x_0)$ on the curve (which can be any rational point, its coordinates need not be positive) and construct a line through that point with rational or infinite slope $s$. There will be at most one second point $(u,x)$ where the line and the hyperbola intersect, and if it exists, the algebraic equations for its coordinates will have degree exactly $1$ over $\mathbb{Q}$, which forces the point to be rational. Thus, the line slope $s$ gets mapped to a rational point $(u,x)$. Conversely, any rational point $(u,x)$ on the curve that is distinct from $(u_0,x_0)$ uniquely defines a line with rational or infinite slope passing through those two points. This correspondence allows parameterization of the rational points on the curve, possibly except the root point $(u_0,v_0)$.

As root point for the parametrization of $\text{(1a)}$ we could use e.g. $$(u_0,x_0) = (y+\frac{1}{2},y-\frac{1}{2})$$ and begin our calculation without thinking further.

However, there is a root point that is much more suitable for the task at hand, although it can be found only in the projective plane: the point at infinity, where the hyperbola in the first (or third) quadrant heads to. Using that point has the consequence that the slope of our intersecting line is fixed to the asymptotic slope $+1$, and that line variation takes the form of a parallel shift now. As you can verify below, the reasoning about parameterizing the rational points still works in this modified setting.

Hence, we use the lines \begin{align} u-x &= s \tag{2a} \\ v-y &= t \tag{2b} \end{align} where $s$ and $t$ are the parameters effecting parallel line shifts. From the constraints $u,v,x,y\in\mathbb{Q}$ we deduce $s,t\in\mathbb{Q}$, and from $u>x$, $v>y$ we find $s>0$, $t>0$. For the intersection of the line $\text{(2a)}$ resp. $\text{(2b)}$ with the hyperbola $\text{(1a)}$ resp. $\text{(1b)}$ in a finite point $(u,x)$ resp. $(v,y)$, we now get \begin{aligned} (2x+s)s = (2u-s)s &= 2y \qquad\therefore& x &= \frac{y}{s}-\frac{s}{2} &u &= \frac{y}{s}+\frac{s}{2} \\ (2y+t)t = (2v-t)t &= 2x \qquad\therefore& y &= \frac{x}{t}-\frac{t}{2} &v &= \frac{x}{t}+\frac{t}{2} \end{aligned} The middle two equations form a linear system for $(x,y)$ parametrized by $(s,t)$: $$\begin{pmatrix}s&-1\\-1&t\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = -\frac{1}{2}\begin{pmatrix}s^2\\t^2\end{pmatrix}\tag{3}$$ Its determinant is $\Delta = st-1$.

Suppose $\Delta=0$, so $st=1$. Then solutions for $(x,y)$ can exist only if $\text{(3)}$ is consistent, that is, $t^2=-s$ and $s^2=-t$, but neither can happen for positive $s,t$. Therefore the system $\text{(3)}$ must be regular, i.e. $\Delta\neq0$, so $st\neq1$. Its solutions must then take the form \begin{align} x(s,t) &= \frac{t^2+s}{2(1-st)}-\frac{s}{2} = \frac{(s^2+t)t}{2(1-st)} & y(s,t) &= \frac{s^2+t}{2(1-st)}-\frac{t}{2} = \frac{(t^2+s)s}{2(1-st)} \tag{4a} \\ u(s,t) &= \frac{t^2+s}{2(1-st)}+\frac{s}{2} & v(s,t) &= \frac{s^2+t}{2(1-st)}+\frac{t}{2} \tag{4b} \end{align} Applying the constraints that $x,y$ shall be positive, we find that $st<1$ must hold in addition to $s>0,t>0$. With these constraints, $\text{(4a)}$ and $\text{(4b)}$ is the rational parameterization of $(x,y)$ resp. $(u,v)$, each restricted to positive entries, in terms of $(s,t)$.

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I know how to do what you did, but I need to use the ancient Greek method, which is what I can't figure out. I can only use one variable and the answer is almost always a pair of fractions. Thanks for the help, though. – Mike Perez Oct 30 '13 at 20:41