Gauss’s Proof by Mathematical Induction

Oswald Baumgart

Chapter Chapter 2 in The Quadratic Reciprocity Law, 2015, pp 7-13 from Springer

Abstract: Abstract 1. Gauss Gauss distinguishes in his first proof, just as Legendre, 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 $$(\frac{q} {p}) = -1$$ .

Keywords: Mathematical Induction; Actual Proof; Coprime; Quadratic Nonresidue; Residue Quality (search for similar items in EconPapers)
Date: 2015
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DOI: 10.1007/978-3-319-16283-6_2

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