# Homework Help: Help with proofs

1. Feb 8, 2005

### Nx2

i just started my second semester with geomtry and am having difficulties with these proofs. i am stuck on this one question which asks:

prove that if n is an odd positive integer, then one of the numbers n+5 or n+7 is dividsible by 4.

so this is what i came up with:

let n = 2k+1

f(n)= n+5
= (2k+1)+5
= 2k+6
= 2(k+3)

So we know that 2 is divisible by 2 and now im guessing i have to prove that (k+3) is divisible by 2 as well. then by using the factor tree thing we can say that since the 2 is divisible by 2 and (k+3) is divisible by 2, f(n) must be divisible by 4, no? but i dont get how to do this... am i doing something wrong?

i did the same exact method with f(n)= n+7 and ended up with f(n)= 2(k+4).

i just dont get all this proving stuff.

Any help would be appreciated, thanks.

- Tu

2. Feb 8, 2005

### dextercioby

You're basically "home",because you have to prove that one of the 2 no. K+3 or K+4 is divisible with 2,thing which is trivial.

Daniel.

3. Feb 8, 2005

### Nx2

so how would i show that (k+3) or (k+4) is divisible by 2... that what i dont understand. it looks to me that not all cases will be divisible by 2 making the statement false, but my teacher says that none of them are false. unless im doing this question totaly wrong... any ideas?

thnx.

- Tu

4. Feb 8, 2005

### dextercioby

Can u show that from two consecutive natural numbers (as is the case with K+3 & K+4),one & only one is divisible by 2...??

Daniel.

5. Feb 8, 2005

### slipF

yea tu it sucks balls (this is phillips)

6. Feb 8, 2005

### Nx2

oooo... omg i cant beleive i didnt see that... so (k+3) and (k+4) are consecutive number, which mean one of them must be even, making one of them divisible by 2 correct?

- Tu

7. Feb 8, 2005

### dextercioby

That's right...

Daniel.

8. Feb 8, 2005

### slipF

yes
also you didnt need to substitute if you think about it, you already had the consecutive pairs

n, n+5, and n+7
n being odd, you are adding 1 to make it even, than another 4
if that isnt divisible, the other pair adds another 2, so it must be divisible if the other wasnt.

9. Feb 8, 2005

### Nx2

Omg man thnx alot... i apreciate it man

- Tu

lol, hey wats up Philips, didnt know u go on here lol

10. Feb 8, 2005

### Nx2

ooo i got yea philips... i should have seen that comming... man i hate this proof stuf... brutal...

11. Feb 8, 2005

### slipF

useful forum
hey daniel while you are here
how would you do

prove that n^5 - 5n^3 + 4n is always divisible by 120 when n is greater than or equal to 3

and prove that there are no integer solutions to the equation 2x + 4y = 5
since both 2 and 4 are even numbers, does that alone prove there are no solutions?
thanks!