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, \(\displaystyle{a}^{{{3}}}\equiv{c}^{{{3}}}{\left({b}\text{mod}{n}\right)}\).

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, \(\displaystyle{a}^{{{3}}}\equiv{c}^{{{3}}}{\left({b}\text{mod}{n}\right)}\).