Find $p,q,r\in \mathbb Q$ such that $r^2-5=p^2$ and $r^2+5=q^2$

Question

Find $p,q,r\in \mathbb Q$ such that $$r^2-5=p^2$$ $$r^2+5=q^2$$ I added and subtracted those two that gave me no luck.

Answer 1

Here is an extension of my comments.

Start from $p^2+q^2=2r^2$. Denote $x=p/r,y=q/r$. Then the equation become $x^2+y^2=2$. We solve all possible rational solutions of the equation $x^2+y^2=2$. It has one solution $P=(1,1)$. Let $Q\ne (1,-1)$ be any other rational point on this circle, then the slope $k$ of $PQ$ is rational. The line equation of $PQ$ can be expressed as $y=k(x-1)+1$. From the line and circle equation one get the $x$-coordinate of $Q$ is $$(1)\quad x_Q=\frac{k^2-2k-1}{k^2+1}.$$

We need to consider another equation $r^2-5=q^2$, which can be rewritten as $$1-\frac{5}{r^2}=x^2.$$ From equation (1), one obtains $$\frac{5}{r^2}=1-x^2=\frac{4k(k-1)(k+1)}{(k^2+1)^2}.$$

Let $Y=(k^2+1)/(2r)$ and $X=k$, the above equation becomes

$$5Y^2=X^3-X, X\ne 0, \pm 1.$$

This equation is equivalent to

$$(2)\quad Y^2=X^3-25 X. X\ne 0, \pm 5.$$

The above elliptic curve has no torsion points other than $X=0,\pm 5$. Since $5$ is a congruent number, the elliptic curve (2) has infinite number solutions. The reference is N.Koblitz' book "Introduction to Elliptic curves and modular forms".

Here is one solution $p=113279/1494696$, $q=4728001/1494696$, $r=3344161/1494696$ with $k=41^2/(5\cdot 12^2)$.

Here are further details on how to find solutions of your problem. This is essentially proposition 1.1 of Koblitz' book. We call an integer $n$ a congruent number if it is area of a right triangle whose sides are rational numbers, i.e., if there exists $X,Y,Z\in \textbf{Q}$ such that $Z^2=X^2+Y^2$ and $n=\frac{1}{2}XY$. We call $(X,Y,Z)$ a solution of the congruent problem for $n$.

We next consider your problem: (Problem (*)) given a positive integer $n$, is it possible to find rational $p,q,r$ such that $r^2-p^2=n, r^2-q^2=-n$.

Proposition 1.1 of Koblitz' book says that the above two problems are equivalent in the following way.

Given a solution $(p,q,r)$ of Problem (*), let $X=q+p,Y=q-p,Z=2r$. Then $(X,Y,Z)$ is a solution of the congruent problem for $n$.

Conversely, given a solution of the $X,Y,Z$ of the congruent problem, one can find a solution of the congruent problem by $r=Z/2,q=(X+Y)/2,p=(X-Y)/2$.

Now your question becomes how to find a congruent problem for 5. Here is one classical solution $(X=20/3,Y=3/2,Z=41/6)$. Using the above correspondence and this classical solution, one can get an easier solution of your problem $$p=31/12,q=49/12,r=41/12.$$

My previous solution is too complicated, and it also comes from Koblitz' book. The elliptic curve method has at least one advantage: from one solution, one can get infinite number solutions by addition law from the elliptic curve.

In general, the congruent number problem is very hard and probably not solved in its most general form. This is an active area of research.

Answer 2

multiplying $$r^2-5=p^2$$ by $-1$ and adding to the second one we obtain $$10=p^2-q^2$$ wrting this equation in the form $$10=(q-p)(q+p)$$ can you finish now?