conditional congruences
Consider congruences^{} (http://planetmath.org/Congruences) of the form
$f(x):={a}_{n}{x}^{n}+{a}_{n1}{x}^{n1}+\mathrm{\dots}+{a}_{0}\equiv \mathrm{\hspace{0.33em}0}\phantom{\rule{veryverythickmathspace}{0ex}}(modm)$  (1) 
where the coefficients ${a}_{i}$ and $m$ are rational integers. Solving the congruence means finding all the integer values of $x$ which satisfy (1).

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If ${a}_{i}\equiv 0\phantom{\rule{veryverythickmathspace}{0ex}}(modm)$ for all $i$’s, the congruence is satisfied by each integer, in which case the congruence is identical (cf. the formal congruence). Therefore one can assume that at least
$${a}_{n}\not\equiv 0\phantom{\rule{veryverythickmathspace}{0ex}}(modm),$$ since one would otherwise have ${a}_{n}{x}^{n}\equiv 0\phantom{\rule{veryverythickmathspace}{0ex}}(modm)$ and the first term could be left out of (1). Now, we say that the degree of the congruence (1) is $n$.

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If $x={x}_{0}$ is a solution of (1) and ${x}_{1}\equiv {x}_{0}\phantom{\rule{veryverythickmathspace}{0ex}}(modm)$, then by the properties of congruences (http://planetmath.org/Congruences),
$$f({x}_{1})\equiv f({x}_{0})\equiv \mathrm{\hspace{0.33em}0}\phantom{\rule{veryverythickmathspace}{0ex}}(modm),$$ and thus also $x={x}_{1}$ is a solution. Therefore, one regards as different roots of a congruence modulo $m$ only such values of $x$ which are incongruent modulo $m$.

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One can think that the congruence (1) has as many roots as is found in a complete residue system^{} modulo $m$.
Title  conditional congruences 
Canonical name  ConditionalCongruences 
Date of creation  20130322 18:52:23 
Last modified on  20130322 18:52:23 
Owner  pahio (2872) 
Last modified by  pahio (2872) 
Numerical id  6 
Author  pahio (2872) 
Entry type  Topic 
Classification  msc 11A07 
Classification  msc 11A05 
Related topic  LinearCongruence 
Related topic  QuadraticCongruence 
Defines  degree of congruence 
Defines  root of congruence 
Defines  root 