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If a and b are integers, is a2−b2 even?
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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?
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28 Sep 2015, 03:29
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 \(a^2b^2\)=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?
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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?
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08 Dec 2016, 03:34
Here is my solution
Given data > a and b are positive integers. Now, a^2b^2 => (a+b)(ab) will be even when a and b are both even or both odd. Note => a+b and ab 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 ab=> odd hence a^2b^2 => odd Sufficient
Statement 2> a=odd If b is odd => a^2b^2=> even If b is even => a^2b^2=> odd
Hence insufficient Hence A
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Re: If a and b are integers, is a2−b2 even?
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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. \(a^2b^2 = (ab)(a+b)\) (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 \(a^2b^2 = (ab)(a+b)\) = odd*odd = even. Sufficient. (2) clearly insufficient.



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Re: If a and b are integers, is a2−b2 even?
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20 Dec 2016, 23:34
a^2b^2= (ab)*(a+b) (1) a^2+2ab+b^2  odd a^2+2ab+b^2=(a+b)^2 (a+b)  odd, thus (ab)  odd, odd*odd=odd, hence a^2b^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?
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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 ab odd also if a is even then b is odd making ab is odd Implies (ab) is odd too hence (ab)(a+b) = Odd * Odd = Odd = a^2b^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?
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21 Mar 2017, 20:17
for B)
a^2 + 2ab + b^2 = odd
notice that 2ab is even >a^2 + b^2 = odd > a^2b^2 = odd
does this work???



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Re: If a and b are integers, is a2−b2 even?
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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. \(a^2−b^2 = (a+b)(ab)\) St 1: \(a^2+2ab+b^2\) is odd or \((a+b)^2\) is odd or a+b is odd. therefore ab 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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