congruence of arbitrary degree
Theorem. A congruence of th degree and modulo a prime number has at most incongruent roots.
Proof. In the case , the assertion turns out from the entry linear congruence. We make the induction hypothesis, that the assertion is true for congruences of degree less than .
We suppose now that the congruence
(1) |
where , has at least incongruent roots . Form the congruence
(2) |
Both sides have the same term of the highest degree, whence they may be cancelled from the congruence and the degree of (2) has a lower degree than . Because (2), however, clearly has incongruent roots , it must by the induction hypothesis be simplifiable to the form and thus be an identical congruence.
Now, if the congruence (1) had an additional incongruent root , i.e. , then the identical congruence (2) would imply
Yet, this is impossible, since no one of the factors (http://planetmath.org/Product) of the left hand side is divisible by . This settles the induction proof.
Cf. http://eom.springer.de/c/c024860.htmSpringerLink.
Example. When , we have
,
,
,
,
,
,
.
Thus only the representants 2 and 4 of a complete residue system modulo 7 (see conditional congruences) are roots of the given congruense. A congruence needs not have the maximal amount of incongruent roots mentionned in the theorem.
References
- 1 K. Väisälä: Lukuteorian ja korkeamman algebran alkeet. Tiedekirjasto No. 17. Kustannusosakeyhtiö Otava, Helsinki (1950).
Title | congruence of arbitrary degree |
---|---|
Canonical name | CongruenceOfArbitraryDegree |
Date of creation | 2013-03-22 18:52:29 |
Last modified on | 2013-03-22 18:52:29 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 10 |
Author | pahio (2872) |
Entry type | Theorem |
Classification | msc 11A05 |
Classification | msc 11A07 |
Related topic | SufficientConditionOfPolynomialCongruence |
Related topic | APolynomialOfDegreeNOverAFieldHasAtMostNRoots |