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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 14 Jun 2017, 04:26
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If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) |xy| is NOT a square of an integer
(2) Point (x, y) is above x-axis



SOLUTION (OE) IS HERE.

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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 23 Jun 2017, 12:05
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OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) |xy| is NOT a square of an integer.

If y = 0 were true, then |xy| would be 0, which is a square of an integer.
If x = -y were true, then |xy| would be y^2, which is a square of an integer (since we are told that y is an integer).

Therefore, since we are told that |xy| is NOT a square of an integer, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.


(2) Point (x, y) is above x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis.
If x = -y were true, then then point (x, y) would be (x, -x), so (non-negative, non-positive), which would mean that it's either on x-axis or below it.

Therefore, since we are told that point (x, y) is above x-axis, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.


Answer: D.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y Updated on: 21 Jun 2017, 03:08
Originally posted by Luckisnoexcuse on 14 Jun 2017, 04:35.
Last edited by Luckisnoexcuse on 21 Jun 2017, 03:08, edited 1 time in total.
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Bunuel
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) |xy| is NOT a square of an integer
(2) Point (x, y) is above x-axis
Show more

P.S. :Second Attempt

Squaring both sides
x^2 - y^2 = x^2 + y^2 +2xy

y(y+x) = 0

So the question basically asks if y=0 OR y= -x

(x^2 - y^2)^(1/2) = x + y


(1) |xy| is NOT a square of an integer
if |xy| is not square of an integer then

|x| is not equal to |y|

Hence y cannot be x,-x but y cannot be 0 as |xy| cannot be 0 (Square of an integer)

Sufficient

(2) Point (x,y) is above x axis shows that y>0 hence y cannot be 0

Also since \(-x \leq y \leq x\), y cannot be equal to -x (since y is positive)

Sufficient

Hence both can solve
Answer D

Please correct me if i am wrong
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 14 Jun 2017, 07:49
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Squaring both sides we get 2y(x+y) = 0 , so the question becomes "Is 2y(x+y) =0 ?" or y =0 or x=-y ?

Statement 1:

|xy| != perfect square ..
means x!=y or x!=-y , but Y could be zero, or non-zero. Not Sufficient

Statement 2 :

Point (x,y) is above X- axis.

Therefore, y>0 and x = anything.

Point could be (-3,3) or ( -4,3)

So, NOT Sufficient .


Answer E


Correct me if i am wrong.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y Updated on: 25 Jun 2017, 23:18
Originally posted by niks18 on 14 Jun 2017, 10:47.
Last edited by niks18 on 25 Jun 2017, 23:18, edited 1 time in total.
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Bunuel
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) |xy| is NOT a square of an integer
(2) Point (x, y) is above x-axis
Show more

simplifying the question stem we get

\(\sqrt{x^2 - y^2} = x + y\) ---- squaring both sides we get

x^2 - y^2 = x^2 + y^2 +2xy, solving this we get

y^2+xy=0 OR y(y+x)=0.

this implies either y = 0 or y = -x

Statement 1: |xy| is not square, which means that y is not equal to -x and y cannot be 0 as in that case |xy| = 0 which is a perfect square. Hence we get a clear "NO" for our question stem. Sufficient

Statement 2: This implies that both x & y are non zero nos and both are above x-axis, so y>0
Also as explained by Bunuel
\(-x \leq x\) so we can say that \(0 \leq 2x\) or \(0 \leq x\)
Hence y =-x is not possible as both x & y are positive. Hence sufficient

Option D
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 14 Jun 2017, 11:46
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There are flaws in all solutions above, even in the one with correct answer.

P.S. Above question and similar question HERE are brand new questions by me and they seem to be quite tricky.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y Updated on: 17 Jun 2017, 03:40
Originally posted by rsgooga on 15 Jun 2017, 06:29.
Last edited by rsgooga on 17 Jun 2017, 03:40, edited 1 time in total.
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\(\sqrt{x^2-y^2}= x + y\) .... Squaring both sides we get

\(x^2-y^2 = x^2+y^2+2xy\)

After simplification we get \(2y(x+y)=0\)
Now we can simply find out that does \(2y(x+y)=0\).
For \(2y(x+y)=0\) to be true either of the following condition must be met:
1. 2y=0
or
2. x+y=0 ---> x=-y (x and y have same value but opposite signs.

Lets Check the statement 1:
This statement tells us that neither x nor y is equal to zero. hence, \(2y(x+y)=0\) may not be valid.
This statement also tells us that x+y is not equal to zero 0. hence, \(2y(x+y)=0\) may not be valid.
Taking both clues together this statement is sufficient to answer the question.

Statement two tells us that:
x,y are above x axis. Well this statement tells us that neither x nor y is equal to zero. hence, \(2y(x+y)=0\) may or may not be valid.
However, there is no clue provided for x+y.
if (2,2) then x+y = 4 and hence, \(2y(x+y)=0\) will not be valid.
But if (-2,2) then x+y= 0 hence, \(2y(x+y)=0\) may be valid.

Thus statement 2 is insufficient
Answer A
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 15 Jun 2017, 07:42
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rsgooga
\(\sqrt{x^2-y^2}= x + y\) .... Squaring both sides we get

\(x^2-y^2 = x^2+y^2+2xy\)

After simplification we get \(2y(x+y)=0\)
Now we can simply find out that does \(2y(x+y)=0\).
For \(2y(x+y)=0\) to be true either of the following condition must be met:
1. 2y=0
or
2. x+y=0 ---> x=-y (x and y have same value but opposite signs.

Lets Check the statement 1:
This statement tells us that neither x nor y is equal to zero. hence, \(2y(x+y)=0\) may not be valid.
This statement also tells us that x+y is not equal to zero 0. hence, \(2y(x+y)=0\) may not be valid.
Taking both clues together this statement is sufficient to answer the question.

Statement two tells us that:
x,y are above x axis. Well this statement tells us that neither x nor y is equal to zero. hence, \(2y(x+y)=0\) may or may not be valid.
However, there is no clue provided for x+y.
if (2,2) then x+y = 4 and hence, \(2y(x+y)=0\) will not be valid.
But if (2,-2) then x+y= 0 hence, \(2y(x+y)=0\) may be valid.

Thus statement 2 is insufficient
Answer A
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Point (2, -2) is NOT above x-axis.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 17 Jun 2017, 03:41
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rsgooga
\(\sqrt{x^2-y^2}= x + y\) .... Squaring both sides we get

\(x^2-y^2 = x^2+y^2+2xy\)

After simplification we get \(2y(x+y)=0\)
Now we can simply find out that does \(2y(x+y)=0\).
For \(2y(x+y)=0\) to be true either of the following condition must be met:
1. 2y=0
or
2. x+y=0 ---> x=-y (x and y have same value but opposite signs.

Lets Check the statement 1:
This statement tells us that neither x nor y is equal to zero. hence, \(2y(x+y)=0\) may not be valid.
This statement also tells us that x+y is not equal to zero 0. hence, \(2y(x+y)=0\) may not be valid.
Taking both clues together this statement is sufficient to answer the question.

Statement two tells us that:
x,y are above x axis. Well this statement tells us that neither x nor y is equal to zero. hence, \(2y(x+y)=0\) may or may not be valid.
However, there is no clue provided for x+y.
if (2,2) then x+y = 4 and hence, \(2y(x+y)=0\) will not be valid.
But if (2,-2) then x+y= 0 hence, \(2y(x+y)=0\) may be valid.

Thus statement 2 is insufficient
Answer A

Point (2, -2) is NOT above x-axis.
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Thanx. My mistake. I corrected now.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 21 Jun 2017, 02:23
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IMO A


question asks if y = 0 or = -x

Statement 1: it is given that x can't be equal to y or - y as the mod is not a square
neither it can be equal to 0 else statement 1 will become invalid.
therefore, both possibilities denied - answer to question is no and statement 1 becomes sufficient

Statement 2: does not give an answer to if y = -x , it is only given that y is not 0

hence insuff

OA: D

Bunuel - please provide detailed solution for the answer when OA is revealed,
This question along with the other similar question (https://gmatclub.com/forum/if-x-and-y-a ... 42602.html) is an absolute nightmare
please provide detailed solution
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IMO A


question asks if y = 0 or = -x

Statement 1: it is given that x can't be equal to y or - y as the mod is not a square
neither it can be equal to 0 else statement 1 will become invalid.
therefore, both possibilities denied - answer to question is no and statement 1 becomes sufficient

Statement 2: does not give an answer to if y = -x , it is only given that y is not 0

hence insuff

OA: D

Bunuel - please provide detailed solution for the answer when OA is revealed,
This question along with the other similar question (https://gmatclub.com/forum/if-x-and-y-a ... 42602.html) is an absolute nightmare
please provide detailed solution
Show more

After the OA is revealed (tomorrow) I'll provide the solutions for both problems.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 21 Jun 2017, 07:37
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Bunuel
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IMO A


question asks if y = 0 or = -x

Statement 1: it is given that x can't be equal to y or - y as the mod is not a square
neither it can be equal to 0 else statement 1 will become invalid.
therefore, both possibilities denied - answer to question is no and statement 1 becomes sufficient

Statement 2: does not give an answer to if y = -x , it is only given that y is not 0

hence insuff

OA: D

Bunuel - please provide detailed solution for the answer when OA is revealed,
This question along with the other similar question (https://gmatclub.com/forum/if-x-and-y-a ... 42602.html) is an absolute nightmare
please provide detailed solution

After the OA is revealed (tomorrow) I'll provide the solutions for both problems.
Show more

Ok thanks. look forward to solutions of both.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 23 Jun 2017, 12:09
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Bunuel
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) |xy| is NOT a square of an integer
(2) Point (x, y) is above x-axis



SOLUTION (OE) IS HERE.
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Check fresh tricky GMAT CLUB question.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 25 Jun 2017, 13:48
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Bunuel

OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) |xy| is NOT a square of an integer.

If y = 0 were true, then |xy| would be 0, which is a square of an integer.
If x = -y were true, then |xy| would be y^2, which is a square of an integer (since we are told that y is an integer).

Therefore, since we are told that |xy| is NOT a square of an integer, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.


(2) Point (x, y) is above x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis.
If x = -y were true, then then point (x, y) would be (x, -x), so (non-negative, non-positive), which would mean that it's either on x-axis or below it.

Therefore, since we are told that point (x, y) is above x-axis, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.


Answer: D.
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Hey Bunuel - can you explain how it is ensured that:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 25 Jun 2017, 20:49
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Bunuel

OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) |xy| is NOT a square of an integer.

If y = 0 were true, then |xy| would be 0, which is a square of an integer.
If x = -y were true, then |xy| would be y^2, which is a square of an integer (since we are told that y is an integer).

Therefore, since we are told that |xy| is NOT a square of an integer, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.


(2) Point (x, y) is above x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis.
If x = -y were true, then then point (x, y) would be (x, -x), so (non-negative, non-positive), which would mean that it's either on x-axis or below it.

Therefore, since we are told that point (x, y) is above x-axis, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.


Answer: D.
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Hi Bunuel

I am bit confused with your justification for statement 2. isn't x=-y same as -x=y (if we multiply both sides of the equation by -1) so in that case (-2,2) is a possibility and the point will be above x-axis?
kindly explain what i am missing here?
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 25 Jun 2017, 22:08
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niks18
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OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) |xy| is NOT a square of an integer.

If y = 0 were true, then |xy| would be 0, which is a square of an integer.
If x = -y were true, then |xy| would be y^2, which is a square of an integer (since we are told that y is an integer).

Therefore, since we are told that |xy| is NOT a square of an integer, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.


(2) Point (x, y) is above x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis.
If x = -y were true, then then point (x, y) would be (x, -x), so (non-negative, non-positive), which would mean that it's either on x-axis or below it.

Therefore, since we are told that point (x, y) is above x-axis, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.


Answer: D.

Hi Bunuel

I am bit confused with your justification for statement 2. isn't x=-y same as -x=y (if we multiply both sides of the equation by -1) so in that case (-2,2) is a possibility and the point will be above x-axis?
kindly explain what i am missing here?
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Please re-read the highlighted part above. x cannot be negative, so this case is not possible.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 25 Jun 2017, 23:21
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niks18
Bunuel

OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) |xy| is NOT a square of an integer.

If y = 0 were true, then |xy| would be 0, which is a square of an integer.
If x = -y were true, then |xy| would be y^2, which is a square of an integer (since we are told that y is an integer).

Therefore, since we are told that |xy| is NOT a square of an integer, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.


(2) Point (x, y) is above x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis.
If x = -y were true, then then point (x, y) would be (x, -x), so (non-negative, non-positive), which would mean that it's either on x-axis or below it.

Therefore, since we are told that point (x, y) is above x-axis, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.


Answer: D.

Hi Bunuel

I am bit confused with your justification for statement 2. isn't x=-y same as -x=y (if we multiply both sides of the equation by -1) so in that case (-2,2) is a possibility and the point will be above x-axis?
kindly explain what i am missing here?

Please re-read the highlighted part above. x cannot be negative, so this case is not possible.
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Thanks Bunuel for highlighting such a critical point. I have edited my earlier post :-D
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 26 Jun 2017, 01:47
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Hello Bunuel.
Your explanation of the question stem is right at all, thank you.
But how did you get the sufficient answer for both statements?

Let's consider the point x=20, y=1 (20, 1).
This point is right for the ststement: 20 and 1 are integers and -20<=1<=20.
Statement 1): |20*1|=20 - is not a square of an integer. But (20^2-1^2)^(1/2)=(399)^(1/2)!=20+1. Insufficient.
Statement 2): Point (20,1) is above x-axis. Just as statement 1: (20^2-1^2)^(1/2)=(399)^(1/2)!=20+1. Insufficient.
Point (20,1) is insufficient for both statements. Hence answer E.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 26 Jun 2017, 01:59
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Hello Bunuel.
Your explanation of the question stem is right at all, thank you.
But how did you get the sufficient answer for both statements?

Let's consider the point x=20, y=1 (20, 1).
This point is right for the ststement: 20 and 1 are integers and -20<=1<=20.
Statement 1): |20*1|=20 - is not a square of an integer. But (20^2-1^2)^(1/2)=(399)^(1/2)!=20+1. Insufficient.
Statement 2): Point (20,1) is above x-axis. Just as statement 1: (20^2-1^2)^(1/2)=(399)^(1/2)!=20+1. Insufficient.
Point (20,1) is insufficient for both statements. Hence answer E.
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If you re-read the solution above carefully you'll see that both statements give you NO answer to the question. Each pair of x and y, satisfying (1) or (2), will give that \(\sqrt{x^2 - y^2} = x + y\) does NOT hold true.
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y 27 Jun 2017, 04:46
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If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) |xy| is NOT a square of an integer
(2) Point (x, y) is above x-axis



SOLUTION (OE) IS HERE.
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√(x^2-y^2) = (x+y)
Only If y=0

Each statement says that y#0
Hence sufficient

Answer option D
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