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a,b,c are consecutire integers where a < b < c. Which

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a,b,c are consecutire integers where a < b < c. Which [#permalink] New post 17 Sep 2007, 12:55
a,b,c are consecutire integers where a < b < c. Which of the following values are not possible for the equation c^2 - b^2 - a^2?

-12
-6
0
3
4

Please provide the most efficient way to answer this and the number theory that applies. Thanks
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 [#permalink] New post 17 Sep 2007, 13:27
I think the only way is to actually calculate it.

(a+2)^2-(a+1)^2-a^2
---> -a^2+2a+3

Only B does not fit.
You can plug in the values for various integers or actually calculate the eqn witth the answer choices and B is the only one that can not be calculated.
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 [#permalink] New post 17 Sep 2007, 13:57
(n+2)^2-(n+1)^2-n^2 = -n^2+2n+3

The question asks for which of the answer choices makes the above expression -n^2+2n+3) not factorable (i.e. prime).

Answer choice B (-6) is the only one that makes the expression prime.

-n^2+2n+3= -6
-n^2+2n+9=0 is not factorable.
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 [#permalink] New post 17 Sep 2007, 22:07
a = n-1
b = n
c = n+1

c^2-b^2-a^2
= (n+1)^2 - n^2 - (n-1)^2
= (n^2 + 2n + 1) - n^2 - (n^2 - 2n + 1)
= 4n - n^2
= n(4-n)

If n = 0, n(4-n) = 0 --> c is out
If n = 1, n(4-n) = 3 --> d is out
If n = 2, n(4-n) = 4 --> e is out
If n = -2, n(4-n) = -12 --> a is out

Ans: B
  [#permalink] 17 Sep 2007, 22:07
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a,b,c are consecutire integers where a < b < c. Which

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