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

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Excellent explanation by Bunuel, but i will try to give some alternative solution. Since the question askes whether x and y are consecutive perfect squares, if yes they must be perfect numbers in all cases, so we can plug some numbers and check. While plugging in our task is to find numbers when x and y are consecutive and when they are not.

Lets take 4, 9, 16 and 25 - consecutive perfect squares.

(1) if x=9, y=4, z=3, then according to the statement we have 9+4=24+1 which is not true, in case of x=16, y=9 and z=4 we have 16-9=8-1 which is again true. Lets try x=25, y=16, z=5 so we have 25-16=10-1, true again. So in all cases have the same conclusions based on the statement. Sufficient - B.

Hope that helps.

(2) if x=9, y=4, z=3, then according to the statement we have 9-4=6-1 which is true, in case of x=16, y=9 and z=4 we have 16-9=+1 which is again not true. When plug the numbers it is advisable to check at least 3 numbers. So lets try x=25, y=16, z=5 so we have 25+16=40+1 - Bingo! In this case the statement supports that X and Y are consecutive. So we have two different conclusions based on the statement. Not sufficient.

Solution is truly great but the way it has been explained is confusing !Also there are typos !

in the highlighted part above we are testing statement 2 so 16-9=8-1 = 7 which is true !



More simply :


1) x + y = 8z +1

Work backwards take values of Z find x, and then find y. ( \(x= z^{2}\hs{3}\) put x and z in statement and find y )

If z=4 ,x= 16 we get y = 17 cannot take this case as condition is that x>y ( for all positive z<5, y>x , hence we cannot take those cases )
If z=5, x=25,y=16, does satisfy, here x and y are consecutive perfect squares
If z=6,x=36, y=13, also does satisfy, here x and y are not consecutive perfect squares.In fact y is not even a perfect square.
If z=7, x=49,y=9 also does satisfy, here x and y are not consecutive perfect squares.

Two cases hence insufficient.



(2) x – y = 2z – 1

Again work backwards take values of Z find x, and then find y. ( \(x= z^{2}\hs{3}\) put x and z in statement and find y )


if Z=1 then x=1 on solving we get y =0 cannot take this case as y is supposed to be positive.

if Z= 2 then x=4 on solving and finding y we get y =1 , here 4 and 1 are consecutive perfect squares
if Z=3 then x=9 then y = 4 , here also x and y are consecutive perfect squares
if z=4 then x= 16 then y = 9 also satisfies the equation and x and y are perfect squares
. ........
Take any positive integer for z and find x and y we will see both x and y are consecutive perfect squares
Lets take z= 10 then x= 100 and we get y =81 here also x and y are consecutive perfect squares,

hence B is sufficient.

Hope this helps.

Can we just consider the consecutive perfect squares as \(n^2\) and \((n-1)^2\) and the difference could then be expressed as \((n^2)\)- \((n^2-2n+1)\)which is the same as 2n-1 which indicates that Statement 2 is sufficient?

Hi Bunuel

How did u get? 5[square_root]x=x----[square_root]x=5?

Pls explain thanks

Reduce by \(\sqrt{x}\):

\(5\sqrt{x}=x\);

\(5\sqrt{x}=(\sqrt{x})^2\);

\(5=\sqrt{x}\).

Hi Bunuel,

How is \sqrt{x}=5 not sufficient?

We did not get that \(\sqrt{x}=5\),the question became: is \(\sqrt{x}=5\)?

It is given that x and y are 2 consecutive squired. This means they are squires of 2 integers whose mod will give 2 consecutive integers.

Let

x = \(n^2\)
y = \((n-1)^2\) As x > y

Option 1

x+y =8z+1 => \(n^2\) + \((n-1)^2\) = 8n+1 => 2\(n^2\) - 2n +1 = 8n +1 => \(2n^2\) - 10n=0 => n=0,5
The equation satisfied for 2 values of n, while for other values of n it do not satisfy.
Hence, INSUFFICIENT

Option 2

x-y = 2n-1 => \(n^2\) - \((n-1)^2\) = 2n-1 => \(n^2\) - (\(n^2\) +1 -2n) = 2n -1 => 2n -1 = 2n -1
Both LHS and RHS are equal. The equation will satisfy for all values of n.
Hence, SUFFICIENT

Answer "B"

In order to determine that x and y are consecutive squares, we need to plug in y=(z-1)^2 with x=z^2 as given. If, in an equation, LHS=RHS, then its Correct.

St.1: x+y=8z+1
z^2+(z-1)^2=8z+1
z^2+z^2+1-2z=8z+1
2z^2-10z=0
Therefore, for LHS=RHS, Z must be 5. (Z cannot be 0, being a positive interger).We are not given value of Z.
Hence, Insufficient

St.2: x-y=2z-1
z^2-(z-1)^2=2z-1
z^2-(z^2+1-2z)=2z-1
2z-1=2z-1
LHS=RHS
Sufficient.

Hence, Ans B

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Bumping for an alternative approach -

Statement I: This statement can be written as

\(z = \sqrt{17-y} + 4\)
When y = 1, x = 64, z = 8 ---> X,Y Not Consecutive Squares
When y = 16, x = 25, z = 5 --> x, y Consecutive squares.

Statement II: This Statement can be written as

\(y = (z-1)^2\) ----> X, Y are Consecutive Squares.

Hence, B.

Hi Bunuel,

Can you help understand in this solution how we deduce Root(y) is an integer.

We start with "If x and y are consecutive squares then Root(x)-Root(y) = 1". We say x and y are consecutive squares to begin with and therefore we form the equation and check with Options 1 and 2.

We know from the question that z=Root(x) and therefore Root(x) is an integer. However for Root(y) we have no such information is given.

Thanks

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