OlmekinjP

2021-01-04

Prove directly from the definition of congruence modulo n that if a,c, and n are integers,$n>1$, and .

### Answer & Explanation

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Step 1
Let a, c and n be integers such that n>1.
We have to prove that if using the definition of congruence modulo n.
Note that, where p and q are integers.
So, by the definition of congruence modulo n, $a\equiv c\left(\text{mod}n\right)⇒t\stackrel{^}{n}\mid \left(a-c\right)$.
If n divides $\left(a-c\right)$, then n divides any integer multiple of $\left(a-c\right).$
Step 2
Since a and c are integers, ${a}^{2}+ac+{c}^{2}$ is also an integer.
Now, n divides any integer multiple of (a−c) implies that n divides $\left(a-c\right)\left({a}^{2}+ac+{c}^{2}\right)$.
But, by the algebraic identity $\left(a-c\right)\left({a}^{2}+ac+{c}^{2}\right)={a}^{3}-{c}^{3}$.
So, we can say that n divides .
Therefore, by the definition of congruence modulo n, ${a}^{3}\equiv {c}^{3}\left(\text{mod}n\right)$.

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