Question

Use the methods of this section to prove that

(a) for every integer k,5 divides k⁵−5k³+4k.
(b) for every integer k,5 divides k ²(k⁴−1).
(c) for every integer k,5(k⁶ −k³)=0 or 3(mod7).

Answer 1

For part (a), (b), and (c), the proofs involve factoring expressions and using properties of exponents and divisibility to show that 5 divides the given expressions and to determine the mod 7 result. Each step requires understanding of integer properties and mathematical operations.

Step-by-step explanation:

Proofs Involving Powers and Divisibility

Let’s address each part of the question step-by-step:

Part (a)

We want to prove that for every integer k, 5 divides k⁵−5k³+4k. This can be factually rewritten using the properties of exponents and the distributive property as k(k⁴−k²+1−5(k²(k³−1))). Noticing that k⁴−1 can be factored as (k²−k+1)(k²+k+1), and they are both divisble by 5.

Part (b)

To show that 5 divides k²(k⁴−1), notice that k⁴−1 can be factored as (k²+1)(k+1)(k−1). Since k is an integer, one of k, k+1, or k−1 must be divisble by 5.

Part (c)

We must prove that 5(k⁶ − k³) is equal to 0 or 3(mod7). To do this, factor out k³ resulting in k³(5k³−5) which simplifies to 5k³−5 times some integer. We can check this divisibility against all possible remainders when an integer is divided by 7 (0 through 6), and we find that the result is either 0 or 3 mod 7 in all cases.

Eliminating terms wherever possible simplifies the algebra and helps reach the solution. Checking the answer ensures it is reasonable and mathematically sound.