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If x, y, and z are positive integers, where x > y and

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If x, y, and z are positive integers, where x > y and [#permalink] New post 15 Nov 2009, 21:47
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If x, y, and z are positive integers, where x > y and z=x^(1/2), are x and y consecutive perfect squares? (A perfect square is defined as the square of an integer. For example, 36 is a perfect square since it equals 6 squared, while 38 is not a perfect square since it is not equal to the square of any integer.)

(1) x + y = 8z +1
(2) x – y = 2z – 1
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 Jun 2013, 04:51, edited 1 time in total.
OA added.
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Re: Are x and y consecutive perfect squares [#permalink] New post 16 Nov 2009, 00:27
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ctrlaltdel wrote:
If x, y, and z are positive integers, where x > y and z=x^(1/2), are x and y consecutive perfect squares? (A perfect square is defined as the square of an integer. For example, 36 is a perfect square since it equals 6 squared, while 38 is not a perfect square since it is not equal to the square of any integer.)

(1) x + y = 8z +1
(2) x – y = 2z – 1

Happy Solving


B

1. x+ y = 8z+1
the values of x and y satisfying this equation is 25 and 16 which are consecutive perfect squares
the values of x and y satisfying this equation is 64 and 1 which are NOT consecutive perfect squares
Hence insuff

2. x -y = 2z-1
the values x and y satisfying this equation will be all consecutive perfect squares (be it [4,1] [9,4][2500,2401]
non consecutive perfect squares will not satisfy the equation
hence suff
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Re: Are x and y consecutive perfect squares [#permalink] New post 16 Nov 2009, 00:42
[Reveal] Spoiler: OA
B
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Re: Are x and y consecutive perfect squares [#permalink] New post 16 Nov 2009, 06:55
What is the source?? This is a heck of a problem. Plugging numbers is painful. Do we have any algebraic approach?
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Re: Are x and y consecutive perfect squares [#permalink] New post 16 Nov 2009, 07:40
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Economist wrote:
What is the source?? This is a heck of a problem. Plugging numbers is painful. Do we have any algebraic approach?


Here you go.
If x and y are consecutive perfect squares, then \sqrt{y} and \sqrt{x} must be consecutive integers.

So, the question: is \sqrt{y}=\sqrt{x}-1?

Square both sides: y=x-2\sqrt{x}+1?

(1) x+y=8z+1
x+y=8\sqrt{x}+1
y=8\sqrt{x}+1-x

So basically question transforms to is 8\sqrt{x}+1-x=x-2\sqrt{x}+1?
10\sqrt{x}=2x?
5\sqrt{x}=x?
\sqrt{x}=5?
x=25?

If x=25 then yes, if not then no. But we don't know the value of x, hence insufficient.

(2) x-y=2\sqrt{x}-1
y=x-2\sqrt{x}+1, which is exactly what we are asked in the question. Hence sufficient.

Answer: B.
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Re: Are x and y consecutive perfect squares [#permalink] New post 16 Nov 2009, 09:08
Thanks Bunuel :) ! Great job..+1K
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Re: Are x and y consecutive perfect squares [#permalink] New post 17 Nov 2009, 19:07
bunuel - sometimes i wonder how easily you solve these questions - you are awesome :)
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Re: If x, y, and z are positive integers, where x > y and [#permalink] New post 26 Sep 2013, 03:00
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Re: If x, y, and z are positive integers, where x > y and [#permalink] New post 11 Mar 2014, 17:31
IM BACK!

The difference of perfect squares can be expressed as an odd number. Or viceversa, an odd number can be expressed as the difference of two consecutive perfect squares

Namely,

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

Hope this clarifies
Re: If x, y, and z are positive integers, where x > y and   [#permalink] 11 Mar 2014, 17:31
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