Gre Math problem?

Question

If n is an od integer, then N^2-1 must be
A.a prime number
B.an odd integer
C.divisible by 8
D.a multiple of 2^n
E a positive integer

please give me some insights how to slove this kind of problem.

Answer 1

See how many you can eliminate by picking an example. Let n be 5, which is an odd integer. 5 squared minus 1 is 24,which is not prime (A), not an odd integer (B), not a multiple of 2^5 (32) (D), so the answer must be C or E.

Can you eliminate one of them easily? Yes - let n equal -1. Then n squared minus 1 is 0, which is not a positive integer. Therefore the answer must be C.

Here's how you can prove the answer. Any odd integer n can be expressed as 2k + 1. (2k+1) squared equals 4k^2 + 4k + 1. Subtract the 1, and you're left with 4k^2 + 4k. Express that as 4k(k+1). Either k or k+1 must be divisible by 2, because of any two consecutive integers, exactly one is divisible by 2, so the expression 4k(k+1) must be divisible by 4*2, or 8.

Answer 2

For each part, you can prove it or give a counter example.

A. it seems wrong (it's obvious). counterexample: 5*5 - 1 isn't prime

E. it's obviously wrong since 1*1 - 1 = 0 is not a pos. int.

B. is obviously wrong

D. i'm not sure what "multiple" means here

THE DIFFICULT ONE is c !!
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C. it can be done by induction. let n >= 3 and n is odd (n=1 holds),

1) n=3, n*n -1 = 8 = 8*1 holds
2) suppose n*n - 1 = 8*m where m is an integer
3) we want to show (n+2)(n+2) - 1 = 8*p

(n+2)(n+2) - 1 = n*n + 4n + 4 -1 = (n*n - 1) + 4n + 4 = 8*m + 4(n+1)

note than n is odd so n+1 is even, so let n+1 = 2*n'

(n+2)(n+2) - 1 = 8*m + 4(n+1) = 8*m + 4(2*n') = 8*m + 8*n' = 8(m + n')

therefore by (n+2)(n+2) - 1 is divisible by 8

by indution we proved n*n - 1 is divisible by 8

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RESULT: always do the obvious parts and then difficult ones. Induction is the easiest way to do these kind of problems involving integers.

Answer 3

Im going to assume that you mean (N^2)-1.

The best way to look at this problem is to get rid of stuff you can disprove
a) start small say 3 3^2-1 = 8, this is not prime

b) using the same example we can see it is not odd

c) This is starting to look like the right answer since any number on the real line is divisible by 8, unless it specifies somewhere you didn't wrtie that it has to divide evenly and output and integer value.

d) Is that n the same as the Big N, if so then when N=5
N^2-1=24 2^5 =32. 24 is not a multiple of 32 this same example works if you can pick any n.

e)It is not always a postive integer because when n=1 n^2-1=0. Which is neither positive or negative.

So i say the best answer is c since you can divide 8 into anything.

Answer 4

n=2k+1
n^2=4k^2+4k+1
n^2-1=4k^2+4k=4k(k+1)

A. not a prime number (divisible by 4)
B. not odd (multiple of 2)
C. divisible by 8 (multiple of 4 and because you have k(k+1) one of these 2 has to be even and therefor multiple of 2)
D.not a multiple of n^2
lets say it were, then n^2-1=k*n^2 (k integer)
n^2=1/(1-k)
but n is also integer, so this wolud be possible for k =0 which means n=1 (which is only a particulary case)
E. it's positive becasue a quadrat is alwais positive (it would be negative for n=0 which is not odd i think)

:)