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Erschienen in:
2015 | OriginalPaper | Buchkapitel
2. Gauss’s Proof by Mathematical Induction
verfasst von : Oswald Baumgart
Erschienen in: The Quadratic Reciprocity Law
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Abstract
1. Gauss
distinguishes in his first proof, just as Legendre, eight different cases according to the different nature of the primes in question, so that the actual proof is seperated into eight proofs. The eight individual cases are:
1.
If \(q = 4n + 1\), \(p = 4n + 1\) and \((\frac{p} {q}) = 1\), then we have to prove that \((\frac{q} {p}) = 1\);
2.
If \(q = 4n + 1\), \(p = 4n + 3\) and \((\frac{p} {q}) = 1\), then we have to prove that \((\frac{q} {p}) = 1\);
3.
If \(q = 4n + 1\), \(p = 4n + 1\) and \((\frac{p} {q}) = -1\), then we have to prove that \((\frac{q} {p}) = -1\);
4.
If \(q = 4n + 1\), \(p = 4n + 3\) and \((\frac{p} {q}) = -1\), then we have to prove that \((\frac{q} {p}) = -1\);
5.
If \(q = 4n + 3\), \(p = 4n + 3\) and \((\frac{p} {q}) = 1\), then we have to prove that \((\frac{q} {p}) = -1\);
6.
If \(q = 4n + 3\), \(p = 4n + 1\) and \((\frac{p} {q}) = 1\), then we have to prove that \((\frac{q} {p}) = 1\);
7.
If \(q = 4n + 3\), \(p = 4n + 3\) and \((\frac{p} {q}) = -1\), then we have to prove that \((\frac{q} {p}) = 1\);
8.
If \(q = 4n + 3\), \(p = 4n + 1\) and \((\frac{p} {q}) = -1\), then we have to prove that \((\frac{q} {p}) = -1\).