If a, b, and c are integers and abc ≠0, is a2 - b2 a multiple of 4?
(1) a = (c - 1)^2
(2) b = c2 - 1
Answer= C
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Potential spoiler below
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I have a question on this problem. If you were to multiply out statement 1, you would get:
a = (c-1)^2
a = (c-1)(c-1)
a = c^2 -2c + 1
So, if you were to plug that in for a in the original equation, you would get:
a^2 - b^2
(a+b)(a-b)
((c^2 -2c + 1) + b)((c^2 -2c + 1) -b)
So, without solving that equation, wouldn't you know that the term has to be divisible by 4 given that the middle term -2c is in there twice so 2c * 2c is 4c^2?
I am probably making the wrong assumption but I feel like that was enough information for me.
(1) a = (c - 1)^2
(2) b = c2 - 1
Answer= C
******
Potential spoiler below
******
******
I have a question on this problem. If you were to multiply out statement 1, you would get:
a = (c-1)^2
a = (c-1)(c-1)
a = c^2 -2c + 1
So, if you were to plug that in for a in the original equation, you would get:
a^2 - b^2
(a+b)(a-b)
((c^2 -2c + 1) + b)((c^2 -2c + 1) -b)
So, without solving that equation, wouldn't you know that the term has to be divisible by 4 given that the middle term -2c is in there twice so 2c * 2c is 4c^2?
I am probably making the wrong assumption but I feel like that was enough information for me.













