Question

For any integer a, prove the following statements:

(a) a2 ≡ 0 or 1 (mod 3)
(b) a3 ≡ 0, 1 or − 1 (mod 7)
(c) a4 ≡ 0 or 1 (mod 5)

Answer 1

For both cases, a⁴ ≡ 0 or 1 (mod 5) holds true for any integer 'a'.

(a) For any integer 'a', a² ≡ 0 or 1 (mod 3).

(b) For any integer 'a', a³ ≡ 0, 1, or −1 (mod 7).

(c) For any integer 'a', a⁴ ≡ 0 or 1 (mod 5).

Step-by-step explanation:

(a) For a² ≡ 0 or 1 (mod 3):

We can consider two cases:

Case 1: If 'a' is divisible by 3, then a = 3k for some integer k. In this case, a² = (3k)² = 9k², which is divisible by 3, hence a² ≡ 0 (mod 3).

Case 2: If 'a' is not divisible by 3, then a = 3k + 1 or a = 3k + 2 for some integer k. In these cases, a² = (3k + 1)² = 9k² + 6k + 1 = 3(3k² + 2k) + 1, which leaves a remainder of 1 when divided by 3. Similarly, for a = 3k + 2, a² ≡ 1 (mod 3).

Therefore, in both cases, a² ≡ 0 or 1 (mod 3) holds true for any integer 'a'.

(b) For a³ ≡ 0, 1, or − 1 (mod 7):

Again, we can consider different cases:

Case 1: If 'a' is divisible by 7, then a = 7k for some integer k. In this case, a³ = (7k)³ = 343k³ = 7(49k³), which is divisible by 7, hence a³ ≡ 0 (mod 7).

Case 2: If 'a' is not divisible by 7, then a can take values of 7k + 1, 7k + 2, 7k + 3, 7k + 4, 7k + 5, or 7k + 6 for some integer k.

Taking each of these cases and calculating the respective values of a³, we find that a³ ≡ 1, −1, 1, −1, 1, −1 (mod 7), respectively.

Therefore, for all the cases, a³ ≡ 0, 1, or − 1 (mod 7) holds true for any integer 'a'.

(c) For a⁴ ≡ 0 or 1 (mod 5):

Here, we again have two cases:

Case 1: If 'a' is divisible by 5, then a = 5k for some integer k. In this case, a⁴ = (5k)⁴ = 625k⁴ = 5(125k⁴), which is divisible by 5. Thus, a⁴ ≡ 0 (mod 5).

Case 2: If 'a' is not divisible by 5, then consider a = 5k + 1, 5k + 2, 5k + 3, or 5k + 4 for some integer k. By calculating the respective values of a⁴, we find that in all cases, a⁴ ≡ 1 (mod 5).

Therefore, we have successfully proven the given statements for any integer 'a'.

Answer 2

To prove the given statements, we consider the remainder when an integer 'a' is divided by 3, 7, or 5, respectively, and then examine the properties of a2, a3, and a4 under modular arithmetic.

Step-by-step explanation:

To prove that for any integer a, the following statements are true:

a2 ≡ 0 or 1 (mod 3)

a3 ≡ 0, 1, or -1 (mod 7)

a4 ≡ 0 or 1 (mod 5)

We will address each statement separately using the concept of modulo arithmetic.

Proof for Statement (a):

Integer a can have one of three possible remainders when divided by 3: 0, 1, or 2. Therefore, we can consider the square of a (a2) modulo 3 in each of these cases:

If a ≡ 0 (mod 3), then a2 ≡ 0² ≡ 0 (mod 3).

If a ≡ 1 (mod 3), then a2 ≡ 1² ≡ 1 (mod 3).

If a ≡ 2 (mod 3), then a2 ≡ 2² ≡ 4 ≡ 1 (mod 3), since 4 is one more than a multiple of 3.

In all cases, a2 is congruent to either 0 or 1 modulo 3, proving statement (a).