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# If a and b are positive integers (a > b), is a^2 - b^2

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Senior Manager
Joined: 05 Oct 2008
Posts: 270

Kudos [?]: 538 [0], given: 22

If a and b are positive integers (a > b), is a^2 - b^2 [#permalink]

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06 Nov 2008, 04:44
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If a and b are positive integers (a > b), is $$a^2 - b^2$$ divisible by 4?

1. a = b + 2
2. a and b are odd

S_1: $$a^2 - b^2 = (b + 2)^2 - b^2 = 4b + 4$$ , which is divisible by 4. S_2: $$a^2 - b^2 = (2n + 1)^2 - (2k + 1)^2 = 4n^2 + 4n - 4k^2 + 4k$$ , which is divisible by 4.

Can someone pls explain how the solution in the explanation is derived? Thanks

Kudos [?]: 538 [0], given: 22

Director
Joined: 14 Aug 2007
Posts: 726

Kudos [?]: 212 [0], given: 0

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06 Nov 2008, 05:36
study wrote:
If a and b are positive integers (a > b), is $$a^2 - b^2$$ divisible by 4?

1. a = b + 2
2. a and b are odd

S_1: $$a^2 - b^2 = (b + 2)^2 - b^2 = 4b + 4$$ , which is divisible by 4. S_2: $$a^2 - b^2 = (2n + 1)^2 - (2k + 1)^2 = 4n^2 + 4n - 4k^2 + 4k$$ , which is divisible by 4.

Can someone pls explain how the solution in the explanation is derived? Thanks

S_1 is pretty obvious.

in S_2 , as any odd number can be expressed as 2*n + 1 (n>=0), it is substituted for 2 different odd numbers (by taking n and k variables)

You can try with substituting numbers as other way to solve this problem.

Kudos [?]: 212 [0], given: 0

Re: No. Properties   [#permalink] 06 Nov 2008, 05:36
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