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Beweis

Es seien ${\displaystyle {}a,b,c}$ die Seiten eines rationalen rechtwinkligen Dreieckes, das den Flächeninhalt ${\displaystyle {}n}$ realisiert, es ist also

${\displaystyle {}a^{2}+b^{2}=c^{2}\,}$

und ${\displaystyle {}{\frac {ab}{2}}=c^{2}}$. Wir setzen

${\displaystyle {}q={\frac {c^{2}}{4}}\,,}$

was ein Quadrat ist. Ferner sind

${\displaystyle {}q-n={\frac {c^{2}}{4}}-{\frac {ab}{2}}={\frac {a^{2}+b^{2}-2ab}{4}}={\frac {(a-b)^{2}}{4}}\,}$

und

${\displaystyle {}q+n={\frac {c^{2}}{4}}+{\frac {ab}{2}}={\frac {a^{2}+b^{2}+2ab}{4}}={\frac {(a+b)^{2}}{4}}\,}$

Es sei umgekehrt ${\displaystyle {}q}$ so, dass ${\displaystyle {}q-n,q,q+n}$ Quadrate sind. Dann sind also ${\displaystyle {}a={\sqrt {q+n}}-{\sqrt {q-n}}}$, ${\displaystyle {}b={\sqrt {q+n}}+{\sqrt {q-n}}}$ und ${\displaystyle {}c=2{\sqrt {q}}}$ 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}}}
{\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}}}