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## Main Question or Discussion Point

**1. Homework Statement**

1. Prove that if n is an even positive integer, then n³-4n is always divisible by 48.

2. Prove taht the square of an odd integer is always of the form 8k+1, where k is an integer.

3. Observe that the last two digits of 7² are 49, the last two digits of 7³ are 43, thel ast two digits of 7^4 are 01, and the last two digits of 7^5 are 07. Prove that the last two digits of 7^201 are 07.

**2. Homework Equations**

Don't know if any are applicable.

**3. The Attempt at a Solution**

Basically the homework handed out today for my semester 2 discrete and geometry class in high school. I have no experience in proofs and the teacher assigned this without doing any examples :(

I need help on my form and revision on how I attempt these proofs and basically what I can work on to make them look more convincing and professional.

1. Let n = 48k, 48k+2, 48k+4

n³-4n divisible by 48

=(48k)³-4(48)

=110592k³-192k

=192k(576k²-1)

Since 192 is divisible by 48, statement is true.

Should I check with 48k+2, 48k+4 afteR?

2. Since k is an odd integer, we can represent it as 2k+1

(2k+1)² = 8k + 1

4k²+4k+1 = 8k +1

4k²+4k=8k

4k(k+1)=8k

8k is divisible by 4k so statement is true.

3. This one i need help the most.

I figured

7^1 = 7

7^2 = 49

7^3 = 343

7^4 = 2401

and it rotates from 7,9,3,1 (last digit) every 4 numbers...so 7^5 would end with a 7.

so 7^4n+1 would yield a number that ends with 7, 7^4n+2 would yield a number that ends with a 9, etc

so if 7^201 ends with a 7

7^4n+1 = 7^201, n must be an integer

4n+1 = 201

4n = 200

n = 50

Conversely for 7^202

7^4n+1 = 7^202

4n+1 = 202

4n = 201

n is not an integer so 7^202 does not end with a 7.

Getting confused...it looks right sometimes but sometimes it doesn't...am I donig the right stuff?