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Lösen einer quadratischen Gleichung im Stile Diophants

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Wir betrachten ein weiteres Problem der Form: Zwei Zahlen finden, deren Summe und deren Summe der Quadrate der Zahlen gegeben sind.

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Summe sei $\\var{sum}$, Quadratsumme $\\var{qsum}$.

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Wir bezeichnen die Differenz der beiden Zahlen mit $2\\cdot x$.

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", "advice": "

a) Es ist $a=\\frac{\\var{sum}}{2}-x=\\var{sum/2}-x$, $b=\\frac{\\var{sum}}{2}+x=\\var{sum/2}+x$.

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b) Es ergibt sich: $\\var{qsum}=2\\cdot\\var{(sum/2)^2}+2x^2\\Leftrightarrow x^2=\\frac{\\var{qsum}-\\var{2*(sum/2)^2}}{2}=\\frac{\\var{(qsum-2*(sum/2)^2)}}{2}=\\var{(qsum-2*(sum/2)^2)/2}\\Leftrightarrow x=\\var{x}$

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Dann ist die kleinere der beiden Zahlen $a=$[[0]] und die größere der beiden Zahlen $b=$[[1]].

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Jeweils einen Term in Abhängigkeit von $x$ eintragen!

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Durch Einsetzen dieser Terme in $\\var{qsum}=a^2+b^2$ ermittelt man daraus:

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$x=$[[2]] , $a=$[[0]] , $b=$[[1]]

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