If a, b, and c are integers and abc ≠ 0, is a^2 – b^2

Question

If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 a multiple of 4?

(1) a = (c – 1)^2

(2) b = c^2 – 1

Need explanation

Bunuel · Accepted Answer

If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 a multiple of 4?

(1) a = (c – 1)^2. This implies that a = c^2 - 2c + 1. However, this is not sufficient, as it provides no information about b.

(2) b = c^2 – 1. Not sufficient, as it provides no information about a.

(1)+(2) From the above, we have:

a^2-b^2 =

=(a-b)(a+b)=

=(c^2-2c+1 -c^2+1)(c^2-2c+1+c^2-1)=

=(2-2c)(2c^2-2c)=

=4(1-c)(c^2-c) .

Therefore, a^2 - b^2 is a multiple of 4. Sufficient.

Answer: C.

Hope it's clear.

BrushMyQuant · Answer

Question can be written as a^2 - b^2 or (a-b)*(a+b) is a multiple of 4
STAT1 alone is not sufficient as we don't know anything about b
STAT2 alone is not sufficient as we don't know anything about a
Taking both together

a = (c-1)^2 and b = c^2 - 1 or b = (c-1)*(c+1)

a+b = (c-1)^2 + (c-1)*(c+1) = (c-1)* (c-1 + c+1) = c*(c-1)
a-b = (c-1)^2 - (c-1)*(c+1) = (c-1)*(c-1 - (c+1)) = (c-1)*(-2) = -2*c*(c-1)
So, a^2 - b^2 = (a-b)*(a+b) = c*(c-1) * (-2*c*(c-1))
= -2*(c^2)*((c-1)^2)

If c is odd then c-1 will be even and c*c-1 will be a multiple of 2
else if c i even then again c*c-1) will be a multiple of 2

So, in both the cases c*(c-1) is a multiple of 2. So, -2*(c^2)*((c-1)^2) will be a multiple of 4 (Actually will be a multiple of 8 too)
So, Answer will be C

Hope it helps!

adkikani · Answer

Bunuel   niks18   chetan2u   amanvermagmat

How about this approach?
If I simplify question stem to best possible, what am I asked?
Is a / b even?

Why?
a. A multiple pf 4 is always even
b. Squaring of odd / even is always odd /even
c. Even - Even = Even.
or Odd - Odd = Even

I did above in my head.
Essentially, either of statements is clearly insuff 
since I need to know both a and b.

What happens when I combine and simplify  a^2  -  b^2 
I get -2c + 1.

Even if I do not know value of c, I know -2 will always be even 
Even + Odd = Odd. So, we do get an unique ans: NO

amanvermagmat · Answer

Hello

Its NOT necessary for a & b to be even for this to be true. a & b could be both odd also. Eg, a=5, b=3, here a^2 - b^2 = 25-9 = 16, which is a multiple of 4.

BrentGMATPrepNow · Answer

Given : a, b, and c are integers and abc ≠ 0

Target question :    Is a² – b² a multiple of 4?

Statement 1 : a = (c – 1)²  
No information about b. NOT SUFFICIENT

If you're not convinced, let's TEST some values.
There are several values of a, b and c that satisfy statement 1. Here are two: 
 Case a : a = 4, b = 2 and c = 3. In this case, a² – b² = 4² – 2² = 12. So, the answer to the target question is  YES, a² – b² IS divisible by 4 
 Case b : a = 4, b = 3 and c = 3. In this case, a² – b² = 4² – 3² = 7. So, the answer to the target question is  NO, a² – b² is NOT divisible by 4 
Since we cannot answer the  target question  with certainty, statement 1 is NOT SUFFICIENT

Statement 2 : b = c² – 1 
No information about a. NOT SUFFICIENT

If you're not convinced, let's TEST some values.
There are several values of a, b and c that satisfy statement 2. Here are two: 
 Case a : a = 10, b = 8 and c = 3. In this case, a² – b² = 10² – 8² = 36. So, the answer to the target question is  YES, a² – b² IS divisible by 4 
 Case b : a = 9, b = 8 and c = 3. In this case, a² – b² = 9² – 8² = 17. So, the answer to the target question is  NO, a² – b² is NOT divisible by 4 
Since we cannot answer the  target question  with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined    
Statement 1 tells us that a = (c – 1)², so a² = (c – 1)⁴
Statement 2 tells us that b = c² – 1, so b² = (c² – 1)²

This means that: a² - b² = (c – 1)⁴ - (c² – 1)²
KEY: the right side is a DIFFERENCE OF SQUARE, which means we can factor it
We get: a² - b² =   
Expand and simplify to get: a² - b² =   
Simplify to get: a² - b² =   
Factor out some 2's to get: a² - b² =   
Simplify to get: a² - b² =   4  (c² - x)(-x + 1)
We can see that a² - b² is clearly divisible by   4  
So, the answer to the target question is  YES, a² – b² IS divisible by 4

Since we can answer the  target question  with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

KarishmaB · Answer

a = (c – 1)^2 = c^2 + 1 - 2c 
 b = c^2 – 1

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

TargetMBA007 · Answer

Bunuel   KarishmaB   BrentGMATPrepNow

I was wondering if you could help me with a fundamental question here. Do we have to assume that a and b would have different values in a question like this, as there seems to be no such restriction in the stem? for e.g. can both a and b be 1?

My guess is Yes, in which case, they still have the same parity, and hence the ans would still be C, as 0 is a multiple of all numbers, but would love to clarify this, and generally what is general assumption about such variables in all DS Questions? (i.e. can they be the same value or not).

Bunuel · Answer

Unless specified otherwise, different variables may represent the same number.

However, in this question, a = b is not possible. a = b would occur when c = 1, leading to a = b = 0. This contradicts the information given in the stem: abc ≠ 0, implying that none of the unknowns can be 0.

If a, b, and c are integers and abc ≠ 0, is a^2 – b^2

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If a, b, and c are integers and abc ≠ 0, is a^2 – b^2

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