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conditional congruences
Consider congruences of the form
$\displaystyle f(x)\;:=\;a_{n}x^{n}+a_{{n1}}x^{{n1}}+\ldots+a_{0}\;\equiv\;0% \;\;(\mathop{{\rm mod}}m)$  (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).

If $a_{i}\equiv 0\;\;(\mathop{{\rm mod}}m)$ 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
$a_{n}\not\equiv 0\;\;(\mathop{{\rm mod}}m),$ since one would otherwise have $a_{n}x^{n}\equiv 0\;\;(\mathop{{\rm mod}}m)$ and the first term could be left out of (1). Now, we say that the degree of the congruence (1) is $n$.

If $x=x_{0}$ is a solution of (1) and $x_{1}\equiv x_{0}\;\;(\mathop{{\rm mod}}m)$, then by the properties of congruences,
$f(x_{1})\;\equiv\;f(x_{0})\;\equiv\;0\;\;(\mathop{{\rm mod}}m),$ 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 $m$.

One can think that the congruence (1) has as many roots as is found in a complete residue system modulo $m$.
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