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

Question

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^2-2z+1 =y
(z-1)^2 = y, z-1 = y^1/2, z=y^1/2+1, since z=X^1/2 , x^1/2 and y^1/2 are consecutive
1. z^2-8z-1 =-y NO solution
Hence B

(1) x + y = 8z +1
(2) x – y = 2z – 1

Bunuel · Accepted Answer

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.

kp1811 · Answer

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     
non consecutive perfect squares will not satisfy the equation
hence suff

ctrlaltdel · Answer

B
Kudos to kp1811

Economist · Answer

What is the source?? This is a heck of a problem. Plugging numbers is painful. Do we have any algebraic approach?

Economist · Answer

Thanks Bunuel

! Great job..+1K

ichha148 · Answer

bunuel - sometimes i wonder how easily you solve these questions - you are awesome

jlgdr · Answer

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

vipulgoel · Answer

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^2-2z+1 =y
(z-1)^2 = y, z-1 = y^1/2, z=y^1/2+1, since z=X^1/2 , x^1/2 and y^1/2 are consecutive
1. z^2-8z-1 =-y NO solution
Hence B

(1) x + y = 8z +1
(2) x – y = 2z – 1

Alt way

2. z^2-2z+1 =y
 (z-1)^2 = y, z-1 = y^1/2, z=y^1/2+1, since z=X^1/2 , x^1/2 and y^1/2 are consecutive
1. z^2-8z-1 =-y NO solution
Hence B

vishwash · Answer

thanks that was a good one... pretty weak in ds got a get the basics first i guess

chandra73 · Answer

So, the question: is  \sqrt{y}=\sqrt{x}-1 ?
Bunnel can you help me on above line. how do you get that from the provided question

Bunuel · Answer

The questions asks: are x and y consecutive perfect squares (given x > y)?

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 ?

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

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