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If a and b are integers, is a2−b2 even?

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If a and b are integers, is a2−b2 even?  [#permalink]

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New post 28 Sep 2015, 03:12
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If a and b are integers, is \(a^2−b^2\) even?

(1) \(a^2+2ab+b^2\) is odd.
(2) a is odd.

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Re: If a and b are integers, is a2−b2 even?  [#permalink]

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New post 28 Sep 2015, 03:29
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reto wrote:
If a and b are integers, is \(a^2−b^2\) even?

(1) \(a^2+2ab+b^2\) is odd.
(2) a is odd.


is \(a^2-b^2\)=even?

Statement 2 is clearly insufficient as you will get different answers with b=odd or b=even.

Statement 1, \(a^2+2ab+b^2\) = odd ---> (a+b)^2 = odd ---> a+b=odd ---> only 2 cases possible : a=odd and b=even or a=even and b=odd. For both the cases you get a "no" for is \(a^2−b^2\) even

A is the correct answer.
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Re: If a and b are integers, is a2−b2 even?  [#permalink]

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New post 18 Nov 2016, 14:02
reto wrote:
If a and b are integers, is \(a^2−b^2\) even?

(1) \(a^2+2ab+b^2\) is odd.
(2) a is odd.


a^2 - b^2 = Even or odd

We know, (Odd)^2 + (Odd)^2 = Even and (Odd)^2 - (Odd)^2 = Even. The same rule applies for Even numbers. Thus, both a and b have to be 'Odd' or 'even'; one being odd makes the sum of difference of the two an odd number.

(1) a^2 +2ab+b^2 = Odd, 2ab will always be even, therefore (a^2 + b^2) meaning is Odd; so, a^2 - b^2 = Odd. -> SUFFICIENT
(2) a is odd. Nothing about b. If is also odd then (a^2 - b^2) is Even, but if b is even, then, (a^2 - b^2) is Odd. - INSUFFICIENT.

.:. IMO, it is 'A'
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Re: If a and b are integers, is a2−b2 even?  [#permalink]

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New post 08 Dec 2016, 03:34
Here is my solution

Given data -->
a and b are positive integers.
Now, a^2-b^2 => (a+b)(a-b) will be even when a and b are both even or both odd.
Note => a+b and a-b will have the same even/odd nature.

Statement 1-->
(a+b)^2=odd
Power does not affect the even odd nature of any integer.
Hence a+b => odd
and a-b=> odd
hence a^2-b^2 => odd
Sufficient

Statement 2-->
a=odd
If b is odd => a^2-b^2=> even
If b is even => a^2-b^2=> odd

Hence insufficient
Hence A

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Re: If a and b are integers, is a2−b2 even?  [#permalink]

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New post 12 Dec 2016, 03:10
reto wrote:
If a and b are integers, is \(a^2−b^2\) even?

(1) \(a^2+2ab+b^2\) is odd.
(2) a is odd.


\(a^2-b^2 = (a-b)(a+b)\)

(1) \(a^2+2ab+b^2 = (a+b)(a+b)\) = odd. Means a+b must be odd as only odd * odd = odd. The sum of two integers is odd if one is odd and other is even. So either a = odd and b = even or b = odd and a = even. In both cases a - b = odd. Now recall \(a^2-b^2 = (a-b)(a+b)\) = odd*odd = even. Sufficient.

(2) clearly insufficient.
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Re: If a and b are integers, is a2−b2 even?  [#permalink]

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New post 20 Dec 2016, 23:34
a^2-b^2= (a-b)*(a+b)
(1)
a^2+2ab+b^2 - odd
a^2+2ab+b^2=(a+b)^2
(a+b) - odd, thus (a-b) - odd, odd*odd=odd, hence a^2-b^2 NEVER can be even SUFFICIENT
(2)
a- odd number, however no information is given about b, INSUFFICIENT

Correct answer is A.
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Re: If a and b are integers, is a2−b2 even?  [#permalink]

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New post 21 Mar 2017, 06:35
Given : a.b are integers

1 ) (a+b)^2 = odd => (a+b) is Odd hence If a is odd b is even making a-b odd also if a is even then b is odd making a-b is odd
Implies (a-b) is odd too

hence (a-b)(a+b) = Odd * Odd = Odd = a^2-b^2

A is Suff

2) A is odd . In suff

Hence answer is A
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Re: If a and b are integers, is a2−b2 even?  [#permalink]

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New post 21 Mar 2017, 20:17
for B)

a^2 + 2ab + b^2 = odd

notice that 2ab is even --->a^2 + b^2 = odd --> a^2-b^2 = odd


does this work???
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Re: If a and b are integers, is a2−b2 even?  [#permalink]

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New post 22 Mar 2017, 02:52
reto wrote:
If a and b are integers, is \(a^2−b^2\) even?

(1) \(a^2+2ab+b^2\) is odd.
(2) a is odd.


\(a^2−b^2 = (a+b)(a-b)\)

St 1: \(a^2+2ab+b^2\) is odd
or \((a+b)^2\) is odd
or a+b is odd. therefore a-b will also be odd

hence \(a^2−b^2\) will be odd. ANSWER

ST 2: a is odd. if b is odd, expression will be even. If b is even, expression will be odd. INSUFFICIENT

Option A
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Re: If a and b are integers, is a2−b2 even? &nbs [#permalink] 22 Mar 2017, 02:52
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If a and b are integers, is a2−b2 even?

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