From Gauss' Disquisitiones Arithmeticae §131:

Sī p est numerus prīmus fōrmae 4n+1, erit +p, sī vērō p fōrmae 4n+3, erit -p residuum vel nōn-residuum cuiusvīs numerī prīmī quī positīvē acceptus ipsīus p est residuum vel nōn-residuum.

(I have taken the liberty to add macrons.)

From what I know about this theorem, that is called quadratic reciprocity nowadays, the translation should roughly be:

If p is a prime number of the form 4n+1, let p* denote +p; otherwise if p is of the form 4n+3, let p* denote -p; then p* is a [quadratic] residue (resp. non-residue) of whichever prime number q that is a [quadratic] residue (resp. non-residue) of this very p.

(I have added p* as an intermediate definition to make the sentence flow better in English.)

The last sentence can also be rephrased as:

p* is a quadratic residue of q iff q is a quadratic residue of p.

However, I do not understand where the phrase positīvē acceptus fits in this translation.


I think the "any other prime" is taken to be positive. I was unable to find an answer by a quick search, but perhaps Gauss is allowing negative prime numbers. Quadratic reciprocity does not work quite the same if you flip signs, and that is why Gauss also used ±p for the first prime.

Most are taught in school that the prime numbers are positive (2, 3, 5, 7, 11…), but it makes sense to regard their negative counterparts (-2, -3, -5, -7, -11…) as primes as well. It depends on how you define a prime; if you only refer to divisibility, then negative numbers make perfectly valid primes. And sometimes this is convenient in modern mathematics, as there are cases where one can speak about divisibility and primes but not positivity.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.