1. The problem statement, all variables and given/known data 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. Relevant 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?