Prove directly from the definition of congruence modulo n that if a,c, and n are integers,n >1, and a -= c (mod n), then a^3 -= c^3(mod n).

Question

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

Answer 1

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

Prove directly from the definition of congruence modulo n that if a,c, and n are integers,n >1, and a -= c (mod n), then a^3 -= c^3(mod n).

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