A solution to a problem of Fermat, on two numbers of which by Euler L.

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Additional resources for A solution to a problem of Fermat, on two numbers of which the sum is a square and the sum of their squares is a biquadrate, inspired by the Illustrious La Grange

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0 So niitzlich Primitivwurzeln zum Rechnen sind, so schwierig ist es, eine Primitivwurzel zu finden. )p"-I Es ist w 2 C == - 1 mod p". )p"-I 2 mod p". Dann ist x eine Primitivwurzel mod p" . Obungsaufgaben 1. Der Kassierer einer Bank stellt fest, daB beim Abpacken der bei ihm im Monat Mai eingezaWten Zehnmarkscheine in Biindeln zu je 2,3, ... , 10 (=: n - 1) jeweils ein Schein, beim Abpacken in Biindeln zu je 11 (=: n) Scheinen keiner iibrigbleibt. a) Wie hoch war die Einnahme der Bank mindestens?

Sei h =y - Xo. Dann ist 'Pp(h) :::; Eo . C. Daher ist ftir aile y E U 'Pp(f(y» = 'Pp(f(xo) + f'(xo) . h + Rz (xo, h) :::; Eo . CZ • Mit demselben Verfahren schatzen wir 'Pp(f'(y» abo R; (X, h) ist das erste RestgJied von f'(X»: ='Pp(f'(xo) + R; (xo, h» =Max {'Pp(f'(xo», 'Pp(R; (xo, h»} = C, da C > 'Pp(R; (xo, h») (:::; 'Pp(h) < C). 'Pp(f'(y» Insbesondere ist f' (y) f 0 fUr aile y E U. iv) Wir definieren den Newton-Operator: T(y) := y - ~(y) f (y) fUr aile y E U. Behauptung: T(y) E U ftir Y E U.

C. Daher ist ftir aile y E U 'Pp(f(y» = 'Pp(f(xo) + f'(xo) . h + Rz (xo, h) :::; Eo . CZ • Mit demselben Verfahren schatzen wir 'Pp(f'(y» abo R; (X, h) ist das erste RestgJied von f'(X»: ='Pp(f'(xo) + R; (xo, h» =Max {'Pp(f'(xo», 'Pp(R; (xo, h»} = C, da C > 'Pp(R; (xo, h») (:::; 'Pp(h) < C). 'Pp(f'(y» Insbesondere ist f' (y) f 0 fUr aile y E U. iv) Wir definieren den Newton-Operator: T(y) := y - ~(y) f (y) fUr aile y E U. Behauptung: T(y) E U ftir Y E U. Beweis: Es ist fey») :::; Max { 'Pp(y -xo), 'Pp (f(Y»)} 'Pp ( y -xo - f'(y) f'(y) :::; Eo' C.

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