Beweis
Es seien
die Seiten eines rationalen rechtwinkligen Dreieckes, das den Flächeninhalt
realisiert, es ist also
-
![{\displaystyle {}a^{2}+b^{2}=c^{2}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09001f97c7393d79ac7cf92d1446e2686af2e727)
und
.
Wir setzen
-
![{\displaystyle {}q={\frac {c^{2}}{4}}\,,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec1deb48fb62f1b7668e52767f50fa6b1ef65b43)
was ein Quadrat ist. Ferner sind
-
![{\displaystyle {}q-n={\frac {c^{2}}{4}}-{\frac {ab}{2}}={\frac {a^{2}+b^{2}-2ab}{4}}={\frac {(a-b)^{2}}{4}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db1e0998099ec28782957758261b00af35eb6c6c)
und
-
![{\displaystyle {}q+n={\frac {c^{2}}{4}}+{\frac {ab}{2}}={\frac {a^{2}+b^{2}+2ab}{4}}={\frac {(a+b)^{2}}{4}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3d683288d1736f55deb4861c9fc400851796ec4)
rationale Quadrate.
Es sei umgekehrt
so, dass
Quadrate sind. Dann sind also
,
und
rationale Zahlen. Diese erfüllen
![{\displaystyle {}{\begin{aligned}a^{2}+b^{2}&=({\sqrt {q+n}}-{\sqrt {q-n}})^{2}+({\sqrt {q+n}}+{\sqrt {q-n}})^{2}\\&=q+n-2{\sqrt {q+n}}{\sqrt {q-n}}+q-n+q+n+2{\sqrt {q+n}}{\sqrt {q-n}}+q-n\\&=4q\\&=c^{2}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a833e634136cad71a343c7b5cace24e721984e3a)
und
![{\displaystyle {}{\begin{aligned}ab&={\left({\sqrt {q+n}}-{\sqrt {q-n}}\right)}{\left({\sqrt {q+n}}+{\sqrt {q-n}}\right)}\\&=q+n-(q-n)\\&=2n.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b58f788e6658c108449bc202ba476c179e551f5)