# Rational solutions of $x^2 - 5 = y^2$, $x^2 + 5 = z^2$

Find all rational solutions $(x,y,z)$ of: $$\begin{cases} x^2 - 5 = y^2 \\ x^2 + 5 = z^2 \end{cases}$$

I was told that this problem is given to Fibonacci by the king at that time. He found one solution is $\big ( \frac{41}{12}, \frac{31}{12}, \frac{49}{12} \big )$.

Can we know how Fibonacci found this solution? Can we actually solve this system of equation?

• I think we should use diophantine equations – Archis Welankar Jan 3 '16 at 16:13
• Yes, I tried to "integer-ize" the problem by setting $x = \frac{a}{b}, y = \frac{c}{d}, z = \frac{e}{f}$, but it looks so messy that way. – SiXUlm Jan 3 '16 at 16:16
• For a small start to a large subject, please see the Wikipedia article on congruent numbers – André Nicolas Jan 3 '16 at 16:21
• This is exercise number $5$ of the first chapter in Koblitz book "Introduction to Elliptic curves and Modular forms", dealing with congruent numbers. – Dietrich Burde Jan 3 '16 at 16:24
• math.stackexchange.com/questions/1584485/… – individ Jan 3 '16 at 16:54

Because $n=5$ is a congruent number, there exists a rational $v=x^2$, such that $v\pm 5$ is a positive rational square. All positive rational solutions are in bijection to the rational sides $X,Y,Z$ of a right triangle with area $5$; in general the correspondence is given as follows (for a squarefree congruent number $n$): $$(X,Y,Z)\mapsto v=(Z/2)^2,\;$$ and conversely $$v\mapsto X:=\sqrt{v+n}-\sqrt{v-n},\; Y:=\sqrt{v+n}+\sqrt{v-n},\;Z:=2\sqrt{v}.$$ In particular, $n$ is congruent if and only if there exists a rational $v=x^2$ such that $v\pm n$ are squares of rational numbers.

To find all rational solutions, a bijection to certain rational points on the elliptic curve $E_n:y^2=x^3-n^2x$ can be used. The elliptic curve $E_5$ for $n=5$ has rank $1$, and all rational points on it are known. Its Mordell-Weil group is given by $E_5(\mathbb{Q})\equiv \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}$, generated by the points $(0,0)$, $(5,0)$ and $(-4,6)$. Take for example $P=(-4,6)$. Then $2P=((\frac{41}{12})^2,-\frac{62279}{1728})$. Let $x=(\frac{41}{12})^2$. Then $x-5=(\frac{31}{12})^2$ and $x+5=(\frac{49}{12})^2$. This gives your first solution $$\frac{41}{12},\; \frac{31}{12},\; \frac{49}{12}.$$ We obtain infinitely many others (all in fact) the same way.

Reference (among many others): K. Conrad's article.

Here's another approach:

$$\frac{{\left( {m}^{4}+{g}^{4}\right) }^{2}}{4\,{g}^{2}\,{m}^{2}}\pm\left( {m}^{2}-{g}^{2}\right) \,\left( {m}^{2}+{g}^{2}\right) =\frac{{\left( {m}^{4}\pm2\,{g}^{2}\,{m}^{2}-{g}^{4}\right) }^{2}}{4\,{g}^{2}\,{m}^{2}}$$

If $$\left( {m}^{2}-{g}^{2}\right) \,\left( {m}^{2}+{g}^{2}\right) =5\,{a}^{2}$$

For example: $$m=3,g=1$$