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If x, y, and z are positive integers, where x > y and
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Updated on: 13 Jun 2013, 05:51
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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
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Originally posted by ctrlaltdel on 15 Nov 2009, 22:47.
Last edited by Bunuel on 13 Jun 2013, 05:51, edited 1 time in total.
OA added.




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Re: Are x and y consecutive perfect squares
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16 Nov 2009, 08:40
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: \(y=x2\sqrt{x}+1\)? (1) \(x+y=8z+1\) \(x+y=8\sqrt{x}+1\) \(y=8\sqrt{x}+1x\) So basically question transforms to is \(8\sqrt{x}+1x=x2\sqrt{x}+1\)? \(10\sqrt{x}=2x\)? \(5\sqrt{x}=x\)? \(\sqrt{x}=5\)? \(x=25\)? If x=25 then yes, if not then no. But we don't know the value of x, hence insufficient. (2) \(xy=2\sqrt{x}1\) \(y=x2\sqrt{x}+1\), 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
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16 Nov 2009, 01:27
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 = 2z1 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
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16 Nov 2009, 01:42



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Re: Are x and y consecutive perfect squares
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16 Nov 2009, 07:55
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
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16 Nov 2009, 10:08
Thanks Bunuel ! Great job..+1K



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Re: Are x and y consecutive perfect squares
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17 Nov 2009, 20:07
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
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11 Mar 2014, 18:31
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



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Re: If x, y, and z are positive integers, where x > y and
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30 Apr 2015, 19:18
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
Alt way
2. z^22z+1 =y (z1)^2 = y, z1 = y^1/2, z=y^1/2+1, since z=X^1/2 , x^1/2 and y^1/2 are consecutive 1. z^28z1 =y NO solution Hence B



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Re: If x, y, and z are positive integers, where x > y and
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30 May 2017, 05:05
thanks that was a good one... pretty weak in ds got a get the basics first i guess



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