Exercise 5.3.12 (Solution not represented by a parametric system)

Question

Show that if $x = \pm (r^2 - 5s^2),\ y = 2rs,\ z = r^2 + 5s^2$ then $x^2 + 5y^2= z^2$. This equation has the solution $x = 2, y = 3, z = 7$. Show that this solution is not given by any rational values of $r,s$.

Richard Ganaye · Accepted Answer

\begin{proof} 
 If $x = \pm (r^2 - 5s^2),\ y = 2rs,\ z = r^2 + 5s^2$, then
 \begin{align*}
 x^2 + 5y^2 - z^2 &= (r^2 - 5s^2)^2 + 20 r^2 s^2 - ( r^2 + 5s^2)^2\
 &=r^4 + 25 s^4 - 10 r^2s^2 + 20 r^2 s^2 - r^4 - 25 s^4 - 10 r^2 s^2\
 &= 0,
 \end{align*}
 thus $x^2 + 5y^2 = z^2$.
 
 (These equations are given by the methods of section 5.6: see p. 250, 251.)
 
 Moreover $(2,3,7)$ is solution of the equation $x^2 + 5y^2= z^2$, since $2^2 + 5\cdot 3^2  = 49 = 7^2$, but $(2,3,7)$ is not given by any rational values of $r,s$.
 
 Assume for the sake of contradiction that
 \begin{align*}
 \left\{
 \begin{array}{ll}
  \pm (r^2 - 5s^2) &= 2\
   2rs&= 3\
   r^2 + 5s^2 &= 7.
 \end{array}
 ight.
 \end{align*}
 The last equation $r^2 + 5s^2 = 7$ has no integer solution, otherwise $r^2 \equiv 2 \pmod 5$, but $2$ is not a square modulo $5$. This shows that the solution $(2,3,7)$ is not represented by the parametric system $x = \pm (r^2 - 5s^2),\ y = 2rs,\ z = r^2 + 5s^2$.
\end{proof}

Exercise 5.3.12 (Solution not represented by a parametric system)

Answers

Comments

Exercise 5.3.12 (Solution not represented by a parametric system)

Answers

Comments

Add answer