Prove directly from the definition of congruence modulo n that if a,c, and n are integers,n >1, and a\equiv c (\bmod n), then a^{3}\equiv c^{3}(\bmod n).

ringearV 2021-04-06 Answered
Prove directly from the definition of congruence modulo n that if a,c, and n are integers,n >1, and \(\displaystyle{a}\equiv{c}{\left({b}\text{mod}{n}\right)}\), then \(\displaystyle{a}^{{{3}}}\equiv{c}^{{{3}}}{\left({b}\text{mod}{n}\right)}\).

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Expert Answer

lobeflepnoumni
Answered 2021-04-08 Author has 25401 answers
Step 1
Let a, c and n be integers such that n>1.
We have to prove that if \(\displaystyle{a}\equiv{c}{\left({b}\text{mod}{n}\right)}\), then \(\displaystyle{a}^{{{3}}}\equiv{c}^{{{3}}}{\left({b}\text{mod}{n}\right)}\) using the definition of congruence modulo n.
Note that, \(\displaystyle{p}\equiv{q}{\left({b}\text{mod}{n}\right)}\) means n|(p-q) where p and q are integers.
So, by the definition of congruence modulo n, \(\displaystyle{a}\equiv{c}{\left({b}\text{mod}{n}\right)}\) implies that n|(a-c).
if n divides (a−c), then n divides any integer multiple of (a−c).
Step 2
Since a and c are integers, \(\displaystyle{a}^{{{2}}}+{a}{c}+{c}^{{{2}}}\) is also an integer.
Now, n divides any integer multiple of (a−c) implies that n divides \(\displaystyle{\left({a}-{c}\right)}{\left({a}^{{{2}}}+{a}{c}+{c}^{{{2}}}\right)}\).
But, by the algebraic identity \(\displaystyle{\left({a}-{c}\right)}{\left({a}^{{{2}}}+{a}{c}+{c}^{{{2}}}\right)}={a}^{{{3}}}-{c}^{{{3}}}\).
So, we can say that n divides \(\displaystyle{\left({a}^{{{3}}}-{c}^{{{3}}}\right)}\) and thus \(\displaystyle{n}{\mid}{\left({a}^{{{3}}}-{c}^{{{3}}}\right)}\).
Therefore, by the definition of congruence modulo n, \(\displaystyle{a}^{{{3}}}\equiv{c}^{{{3}}}{\left({b}\text{mod}{n}\right)}\).
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