Prove directly from the definition of congruence modulo n that if a,c, and n are integers,n&gt;1, and a≡c(mod n),then a3≡c3(mod n).

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Prove directly from the definition of congruence modulo n that if a,c, and n are integers,n&amp;gt;1, and a≡c(mod n),then a3≡c3(mod n).

tafzijdeq · Accepted Answer

Step 1 Let a, c and n be integers such that n&amp;gt;1. We have to prove that if a≡c(modn),then a3≡c3(modn) using the definition of congruence modulo n. Note that, p≡q(modn)means n∣(p−q) where p and q are integers. So, by the definition of congruence modulo n, a≡c(modn)⇒tn^∣(a−c). If n divides (a−c), then n divides any integer multiple of (a−c). Step 2 Since a and c are integers, a2+ac+c2 is also an integer. Now, n divides any integer multiple of (a−c) implies that n divides (a−c)(a2+ac+c2). But, by the algebraic identity (a−c)(a2+ac+c2)=a3−c3. So, we can say that n divides (a3−c3)andthus n∣(a3−c3). Therefore, by the definition of congruence modulo n, a3≡c3(modn).

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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