Question 527290: Prove : If m is an odd integer, then 4 divides m^2 +2m+5
I have gotten this so far but i dont know if its right and how to finish working it out!
Proof: Let m be an odd integer
there is an integer such that m=2k+1
then m^2 +2m +5 = (2k+1)^2 + 2(2k+1) +5
then m^2 + 2m +5= (4k^2 +4k +1) + (4k+2) +5
then m^2 +2m +5= (4k^2 +4k+4k) +1+2+5
then m^2 +2m +5 = 4(k^2 +k+k) + 8
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! You're correct. All you need to do now is factor it to 4(k^2 + 2k + 8). Since k^2 + 2k + 8 is an integer, the expression is divisible by 4.
Another way to prove it is to use modular arithmetic, or arithmetic dealing with remainders. If m is an odd integer, then m ≡ 1 mod 2 (i.e. m divided by 2 leaves a remainder 1. The ≡ sign means "equivalent"). Then m^2 ≡ 1 (mod) 4, 2m ≡ 2 (mod 4), and 5 ≡ 1 (mod 4). We can add modulos or residues the same way we add numbers; in this case, m^2 + 2m + 5 ≡ 1 + 2 + 1 ≡ 4 ≡ 0 (mod 4). Note that if a number is m mod n, we can subtract any multiple of n and still obtain an equivalent result. Since the expression is 0 mod 4, it is divisible by 4 (since it leaves remainder 0).
|
|
|
