Exercise 5.4.17 (Fermat's alternative form)

Question

The preceding problem was asked by Fermat in the following alternative form. If the lengths of the sides of a right triangle are rational numbers, then the area of the triangle cannot be the square of a rational number. Derive this from the former version.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} 
Assume for the sake of contradiction that there are positive rational numbers $(x,y,z,t) \in (\Q^*_+)^4$ such that
\begin{align}
\left\{
\begin{array}{rl}
x^2 + y^2 &= z^2\
\frac{1}{2}\, xy &= t^2.
\end{array}
ight.
\end{align}
Let $d$ be a common denominator of $x, y, z$, so that
$$x = \frac{p}{d},\qquad y = \frac{q}{d},\qquad z = \frac{r}{d},\qquad t = \frac{m}{n},$$
where $x,y,z,d,m,n$ are positive integers. Then (1) implies
\begin{align*}
\left\{
\begin{array}{rl}
p^2 + q^2 &= r^2,\
\frac{1}{2} \, pq &= \left(\frac{md}{n} ight)^2.
\end{array}
ight.
\end{align*}
We may write $\frac{md}{n} = \frac{u}{v}$, where $ \frac{u}{v}$ is a reduced fraction, so that $u\wedge v = 1$ and $v>0$. Then $pq v^2 = 2 u^2$. Since $v^2 \mid 2 u^2$, and $v^2 \wedge u^2 = 1$, we obtain $v^2 \mid 2$, therefore $v^2 = 1$ or $v^2 = 2$. But $v^2 = 2$ has no solution, thus $v^2 = 1,$ and $v = 1$. This gives
\begin{align}
\left\{
\begin{array}{rl}
p^2 + q^2 &= r^2\
\frac{1}{2}\, pq &= u^2,
\end{array}
ight.
\end{align}
where $p,q$ and $u$ are integers. But we have proved in Problem 16 that this is impossible.

If the lengths of the sides of a right triangle are rational numbers, then the area of the triangle cannot be the square of a rational number. 
\end{proof}

Exercise 5.4.17 (Fermat's alternative form)

Answers

Comments

Exercise 5.4.17 (Fermat's alternative form)

Answers

Comments

Add answer