For how many ordered pairs (x , y) that are solutions of the equation

(x−y)^2+2y^2=18 - Includes both +ve and -ve Integers

(x−y)^2 can have Perfect squares values of 0,1,4,9,16 only.
If (x−y)^2 is odd then 2*y^2 must be odd - which is not possible. So (x−y)^2 can take only 0,4 or 16.

If (x−y)^2 is 4 then 2*y^^2=14; y^2=7 not possible.
Hence (x-y)^2 can be 16 which gives y=1 or -1. If Y=1 then X=5 or -3 and Y=-1 then X= 3 or -5.
When (x-y)^2 is 0, 2y^=18, y=3,-3. Then x=3,-3
Hence 4 ordered pairs are possible (5,1),(-3,1),(3,-1) , (-5,-1), (3,3) and (-3,-3).

IMO C

(x-y)^2 and 2y^2 are always non negative

So let's see what are the cases we can get a perfect square + 2*perfect square = 18 (We take perfect squares since we need x and y to be integers)

Perfect squares: 0, 1, 4, 9, 16 ------- (1)
2*Perfect squares: 0, 2, 8, 18 ------- (2)

So only cases when an element from list 1 and an element from list 2 add up to 18 is (0+18) and (16+2)

(0+18) - Possible when (x,y) = (3,3) and (-3,-3) - 2 cases
(16+2) - Possible when (x,y) = (5,1), (-5,-1), (-3,1) and (3,-1) - 4 cases

Therefore we have 6 cases in total

Answer is (C)

(x-y)^2+2y^2=18

As \(18=0^2+2*3^2\)
and \(18= 4^2+2*1^2\)

Hence there are 2 cases possible

Case 1- when \((x-y)^2\)=0 and \(y^2\)=9
\((x-y)^2=0\)
or x-y=0.....(1)

\(y^2=9\)
y= 3 or -3

We have 2 distinct integral values of y and 1 integral value of x-y.
Hence number of solutions possible= 2*1=2



Case 2 - When \((x-y)^2=16\) and \(y^2=1\)
\((x-y)^2=16\)
(x-y)=4 or -4

and \(y^2=1\)
y=1 or -1

We have 2 distinct integral values of x-y and 2 distinct integral values of y
Number of solutions possible= 2*2=4

Total number of solutions possible= 2+4=6

IMO C

[There are couple of reasons for why we don't need to find the solutions of (x, y)
1. If (x-y) and y are integers, x will always be an integer.
2. Values of y are different in both cases, hence there is no overlap possible.
3 it will save some precious time. ]

Given,

(x − y)^2 + 2*y^2 = 18

This can be seen as 18 being the sum of 2 squares.

Case 1:

=> The first perfect square less than 18 is 16. Let us see if we can have integer values of x and y such that one of the squars is 16

if y = 1, we can try to get (x - y)^2 to be 16.

=> x = 5, y = 1 satisfies the equation. Now let us play with the signs of x and y and see which combinations satisfy the equation. By doing so we get the following 3 combinations of x and y.

1. (5 , 1)
2. (-5 , 1)
3. (-5 , -1)

Case 2:

=> The second perfect square less than 18 is 9. Let us see if we can have integer values of x and y such that one of the squars is 9

if y = 3, we can try to get (x - y)^2 to be 9.

=> x = 3, y = 3 satisfies the equation. Now let us play with the signs of x and y and see which combinations satisfy the equation. By doing so we get the following 3 combinations of x and y.

4. (3 , 3)
5. (-3 , -3)
6. (-3 , 3)

Case 3:

=> The third perfect square less than 18 is 4. Let us see if we can have integer values of x and y such that one of the squars is 4

By putting multiple values of y we see that we do not have an integer pair for x and y such that the equation satisfies where one of the squares in the LHS is the perfect square 4.

Thus we conclude that we have 6 unique integer pairs of x and y which satisfy the given equation.

Answer: C

(x−y)^2+2y^2=18

For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

(x-y)^2 + 2y^2 = 18
In the above equation both square terms will be positive and less than 18 respectively.

By trail and error,
If y = 0, then solving we get x^2 = 18.. not a perfect square.
If y = 1, then solving we get (x-1)^2 = 16 --> x-1 = (+/-)4 --> x = 5 or -3. Hence two pairs (5,1) and (-3,1)
If y = 2, then solving we get (x-2)^2 = 10...not a perfect square.
If y = 3, then solving we get (x-3)^2 = 0 --> x = 3. One pair (3,3)
If y = 4, then 2y^2 becomes > 18 hence y cannot take any value > 3

Similarly y can take negative numbers -1, -2, -3... anything -4 and below we will have second term > 18 hence not possible.

If y = -1, then x+1 = (+/-)4 --> x = 3, -5. Hence two pairs (3, -1) and (-5, -1)
If y = -2, then (x+2)^2 = 10. Not a perfect square.
If y = -3, then (x+3)^2 = 0 --> x = -3. One pair (-3, -3)

Hence total pairs (x, y) are (5,1); (-3,1); (3,3); (3, -1); (-5, -1); (-3, -3)
Total number of pairs = 6.

A. 2
B. 4
C. 6
D. 8
E. 10

Answer Choice: C

We need to find ordered pairs such that x and y both are integers.
Let us look at the equation once and see what we can determine:

\((x – y)^2\) will always be positive -> (a)
\(y^2\) will always be positive -> (b)
Thus, 18 will be the result of the sum of squares of two positive numbers. -> (c)

Looking at the equation and from (b) and (c) we find that y can take a range of value such that \(-3 \leq{y} \leq{+3}\)
Thus, we can find all ordered pairs in the given range of y

y = -3
\((x + 3)^2\) – 18 = 0
x = 3
So (3, -3) is a ordered pair. -> [1]

y = -2
\((x + 2)^2\)+ 8 =18
\((x + 2)^2\) = 10 -> 10 is not a square of any integer.

y = -1
\((x + 1)^2\) – 2 = 18
\((x + 1)^2\) = 20 -> 20 is not a square of any integer.

y = 0
\(x^2\) = 18
x = -9 OR x = 9
So (-9, 0) is a ordered pair. -> [2]
So (9, 0) is a ordered pair. -> [3]

y = 1
\((x – 1)^2\) + 2 = 18
\((x – 1)^2\) = 16
x = 5 OR x = -3
So (-3, 1) is a ordered pair. -> [4]
So (5, 1) is a ordered pair. -> [5]

y = 2
\((x – 2)^2\) = 10 -> 10 is not a square of any integer.

y = 3
\((x -3)^2\) + 18 = 18
x = 3
So (3, 3) is a ordered pair. -> [6]

From [1], [2], [3], [4], [5] and [6] we have 6 ordered pairs.

Answer C

Given: (x-y)^2 +2y^2 = 18
To Find: No. of pairs of (x,y) for which x and y will be integers.

(x-y)^2 +2y^2 = 18 can be written as (x-y)^2 = 18 - 2y^2
=> (x-y)^2 = 2(9-y^2)
=> (x-y)^2 = 2 (3+y)(3-y)
Now, since L.H.S is a perfect square, y can take values between [-3,3] only.
(Any value greater than 3 or less than -3 will make R.H.S negative and L.H.S can't be negative as it's a perfect square.)
i.e. -3,-2,-1,0,1,2,3
For y = +3 or -3, R.H.S will be 0
=> x = +3 or -3
=> 2 pairs (3,3) and (-3,-3)
For y = +2 or -2,R.H.S will be 10
Now, 10 is not a perfect square so we will not get integral solutions for x.
For y = +1 or -1,R.H.S will be 16
Now, 16 is a perfect square
=> (x+1)^2 = 16 and (x-1)^2 = 16
=> x+1 = +-4 and x-1 = +-4
we will get 4 pairs by the above 2 quadratics i.e. (5,1),(-3,1),(3,-1),(-5,-1)
For y = 0,R.H.S will be 18
Now, 18 is not a perfect square so we will not get integral solutions for x.
=> total 6 integral solutions i.e. (3,3),(-3,-3),(5,1),(-3,1),(3,-1),(-5,-1)
Option C.

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