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# If x > 1 is (x+y)(x-y) an even integer?

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If x > 1 is (x+y)(x-y) an even integer?  [#permalink]

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08 Apr 2015, 00:33
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If x > 1 is (x+y)(x-y) an even integer?
1) x+y is even
2) 2x = prime

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Re: If x > 1 is (x+y)(x-y) an even integer?  [#permalink]

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08 Apr 2015, 01:16
1
1
#1:
1)$$x = 5, y = 5$$
$$x + y = 10$$, even
$$(x+y)*(x-y) = 0$$, even integer
2)$$x = 5 + \sqrt{2}, y = 5 - \sqrt{2}$$
$$(x+y)*(x-y) = 10*2*\sqrt{2}$$, not even integer
#1 - insufficient

#2:
Since X is bigger than 1 the only way we can get prime number as $$2*x$$ is if X is pretty much a half of prime number, for example x = 3,5 (7), x =2,5(5),x = 8,5 (17) etc. Its worth mentioning that the number will always look like "a,5" with 5 being the only possible decimal and a - combination of units and tenth digits if necessary.
Now, with just #2 we have no idea about value of y, so with x = 3,5 we can either choose y = 3,5 which would give us 0, even integer, or y = 3, which would give us $$6,5*0,5 = 3,25$$ - not even integer
#2 - insufficient

#1 + #2.
Using the information we accumulated from #2 and taking into account #1 we can conclude that y is also a number which looks like "b,5" with 5 being the only possible decimal. In this case whether you subtract y from x or add y to x you will get an integer value. "a,5" - "b,5" = "a-b" and "a,5" + "b,5" = "a + b + 1". For instance, x = 5,5, y = 3,5: (x+y)*(x-y) = 9*2 = 18. To figure out if we get only even integer as a result lets generalise our numbers:
x = a + 0,5
y = b + 0,5
a, b - integers bigger than 1
(x+y)*(x-y) = (a+b+1)*(a-b)
1)a - even, b - odd: even*odd = even - good
2)a - even, b - even: odd*even = even - good
3)a - odd, b - odd: odd*even = even - good
That makes #1 and #2 combined sufficient to answer our question, thus answer C
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Re: If x > 1 is (x+y)(x-y) an even integer?  [#permalink]

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09 Apr 2015, 00:15
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Harley1980 wrote:
If x > 1 is (x+y)(x-y) an even integer?
1) x+y is even
2) 2x = prime

Question: Is (x+y)(x-y) an even integer?

1) x+y is even

x = 4, y = 2, x+y = 6 (even integer), x-y = 2. So (x+y)(x-y) = 12, an even integer
x = 7/3, y = 5/3, x+y = 4 (even integer), x - y = 2/3 (not an integer so (x+y)(x-y) is not an integer)
Not Sufficient.

2) 2x = prime
Since 2x is prime, if it were 2, x would be 1 but we are given that x > 1.
So 2x must be an odd prime number. So x = Odd prime/2 such as 5/2, 7/2, 11/2 etc.
y can be anything so we cannot say that (x+y)(x-y) is an even integer.

Using both, x + y = Even integer
Odd prime/2 + y = Even integer
So y must be an Odd integer/2.
This means x - y = Odd integer/2 - Odd integer/2 = Even integer/2 = Integer

So (x+y)*(x-y) = Even integer * Integer = Even Integer
Sufficient.

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Re: If x > 1 is (x+y)(x-y) an even integer?  [#permalink]

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21 Nov 2016, 06:30
1
Top Quality Even/Odd question.

Here we need to find whether x^2-y^2 is even or not
Statement 1
Taking test cases approach
x=4
y=2
x^2-y^ will be even
x=4.8
y=1.2
x^2-y^2 will never be an integer. So it is neither even nor odd
Hence not sufficient

Statement 2
2x=prime
Hence x=prime/2
note that x>1
so the prime in question here can never be 2
hence x=> odd prime/2
Again making test cases
x=1.5
y=0.5
x^2-y^2=> even
x=1.5
y=0.99
x^2-y^2 will not be integer
Hence Insufficient

Combing the two statements here
y must be off the form something.5 so that for x (which itself is something.5) => x+y = even
Also note that x-y will be an integer(both number have a 5 after decimal)
hence x+y and x-y will be integers
Hence x^y-y^2 must be even

Hence C
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Re: If x > 1 is (x+y)(x-y) an even integer?  [#permalink]

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21 Feb 2018, 08:42
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Re: If x > 1 is (x+y)(x-y) an even integer?   [#permalink] 21 Feb 2018, 08:42
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