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

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Joined: 19 Oct 2018
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For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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Updated on: 16 Jul 2019, 13:02
1
(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. ]

Originally posted by nick1816 on 16 Jul 2019, 10:33.
Last edited by nick1816 on 16 Jul 2019, 13:02, edited 1 time in total.
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For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 10:40
Quote:
(x−y)2+2y2=18(x−y)2+2y2=18

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

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

18 can be a sum of 2 terms wherein one is a square ($$(x-y)^2$$) and another is a multiple of 2 (2* $$y^2$$)

Only possible pair is 16 + 2 = 18

Now either $$(x-y)^2 =16$$ and $$2y^2 = 2$$ or $$(x-y)^2$$= 2 and $$2y^2 = 16$$

As for $$2 y^2= 16$$, y wont be an integer so we rule out scenario 2.
Hence for first scenario, $$(x-y)^2 =16$$ and $$2y^2 = 2$$
$$2y^2 = 2$$
therefore y = +1,-1
and for $$(x-y)^2 =16$$
(x-y) = +4, - 4
Therefore each will have 2 pairs of values (One with y=1 and another with y=-1). So total 4 pairs.
Option B
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 10:51
(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?

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

If we look at the equation it tells us that (X-Y)^2 will always be integer and that too of an even number,

since (x-y)^2 = 18 - 2y^2->>> even
the even square less than 18 are only 16 and 4 so effectively 2 pairs can help us to reach 18 on RHS,

y = 1, x = 5
y=-1 x= 3

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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 11:07
(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?

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

(x−y)^2+2y^2=18 or, (x−y)^2 =18 - 2y^2
it shows that y can not be more than 3 as (x−y)^2 will be positive. so, there are 6 values of y are possible (-3, -2, -1, 0, 1, 2, 3).
So, plugging in the value of y, the possible integer value of x and y are: (-3,-3 ), (-5, -1), (5, 1), (3, 3)

So, the correct answer choice is (B)
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 11:42
1
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.

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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 11:46
1
(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

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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 12:09
x−y
)
2
+2
y
2
=18
(x−y)2+2y2=18

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

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

Solution:

This problem can be approached by substituting the values of x and y to solve for 18.

$$(x-y)^2$$+2$$y^2$$=18
Substittuing ,
x=5 and y=1

$$(5-1)^2$$+2*$$1^2$$=$$4^2$$+2=18

x=-3 and y=1
$$(-3-1)^2$$+2*$$1^2$$=(-4)^2+2=18

Although there are other values of x and y which satisfy this equation but the value of ordered pair is not integer(for eg when x=0 then y=+$$\sqrt{(6)^}$$ and -$$\sqrt{(6)^}$$.But this value of y is not an integer..Hence we have two ordered pairs of (x,y) i.e (5,1) and (-3,-1) which are the solution of the equation.

Hence A IMO
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 12:15
1
$$(x−y)^2+2y^2=18$$

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

Another question that requires trial and error!! Whenever I see trial and error questions, I feel there must be more elegant solutions that I am missing. So I will look forward to the elegant solutions tomorrow!

Here is how I solved it:

We know that x and y are both integers...

Let $$(x-y)^2$$ = A .... Here, A is an integer because x and y are integers. Also A is non negative because A is a square of integer.
Let $$2y^2$$ = B.... Here, B is integer because y is integer. Also, B is non negative because B is 2 times square of integer.

A + B = 18

We need to find combinations of Non negative integers A and B such that A+B = 18

A and B will be less than or equal to 18 because both are non negative and A has to be a PERFECT SQUARE
Also, B/2 has to be a perfect square

Below are all possibilities with A as perfect square
0+18 = 18 .... This is possible because 18/2 is perfect square
1+17 = 18..... NOT POSSIBLE.... 17/2 is not perfect square
4+14 = 18..... NOT POSSIBLE.... 14/2 is not perfect square
9+9 = 18..... NOT POSSIBLE.... 9/2 is not perfect square

16+2 = 18..... This is possible because 2/2 is perfect square

Possible combinations of A and B are:
A = 0 and B = 18.... I
A = 16 and B = 2..... II

I
A=0
$$(x-y)^2$$ = 0
x-y = 0
x=y

B=18
$$2y^2=18$$
y=+ or - 3

If y = 3, x = 3....(1)
If y = -3, x=-3......(2)

II
A=16
$$(x-y)^2=16$$
x-y=+ or -4
x=y+4 or x = y-4

B=2
$$2y^2=2$$
y=+ or - 1

If y=1,
x = 5 .... (3)
or
x = -3.... (4)

If y = -1,
x = 3... (5)
or
x = -5.... (6)

Answer: C - 6 possible pairs of (x,y) are solutions to above equations.
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 12:48
1
$$(x−y)^2$$+2$$y^2$$ = 18 ---(1)

$$(x−y)^2$$ = 18-2$$y^2$$

since $$(x−y)^2$$ ≥ 0
so 18-2$$y^2$$≥ 0
9-$$y^2$$≥ 0
(y+3)(y-3)≤ 0

values of y---- -3≤ y ≤3---(-3,-2,-1,0,1,2,3)

when y=-3
solving the equation -(1) we get x=-3
we get (-3,-3)

when y=-2 we will not get the integer value of x (given both x and y are integers)
so y=-2 is not possible

when y=-1
solving equation -(1) we get
$$(x+1)^2$$ = 16
from here we get x=3 and x=-5 (both values of x satify equation -(1) for y=-1)
so we get (3,-1) and (-5,-1)

when y=0 we will not get the integer value of x so y=0 is not possible

when y=1
solving equation -(1) we get
$$(x-1)^2$$ = 16
from here we get x=5 and x=-3 (both values of x satify equation -(1) for y= 1)
so we get (5,1) and (-3,1)

y=2 is not possible since we will not get the integer value of x

when y=3 we get x=3 solving equation --(1)
so (3,3)

we get the total ordered pair (x,y) = (-3,-3), (3,-1), (-5,-1), (5,1), (-3,1), (3,3)
total 6 ordered pair
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 12:59
For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

First of all, we need to simplify the equation
(x-y)^2 +2y^2 = 18
(x-y)^2 = 2*9 - 2y^2
(x-y)^2 = 2*(9-y^2)

From the given equation we see that (9-y^2) after multiplying by 2 must equals something squared (= (x-y)^2).
We need y and x to be integers, so 2*(9-y^2) may equals 0, 1, 4, 9, 16, 25, 36 etc.
From 2*(9-y^2) = 0 and (x-y)^2 = 2*(9-y^2) we have y = -3, x = -3 OR y = 3, x = 3
From 2*(9-y^2) = 1 we can't have integer y
From 2*(9-y^2) = 2 we can't have integer y
From 2*(9-y^2) = 4 we can't have integer y
From 2*(9-y^2) = 9 we can't have integer y
From 2*(9-y^2) = 16 and (x-y)^2 = 2*(9-y^2) we have y = 1 and x = 5 OR y = -1 and x = 3

we will not consider every squared integer number more than 16 (for example, 2*(9-y^2) = 25), because 2*(9-y^2) can't be more than 18.

So the answer is 4 pairs - B
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 13:15
(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?

We need sum of both (x−y)^2 and 2y^2 to be 18.
As both are squares 18 is sum of two positive integers.

If we consider y to be 1.
then 2y^2 = 2
therefore, (x−y)^2 must be 16
x-y = 4
y is 1 so x will be 5
( 5, 1 )

We can take y as
1,
$$\sqrt{2}$$
$$\sqrt{3}$$
$$\sqrt{4}$$
$$\sqrt{5}$$
$$\sqrt{6}$$
$$\sqrt{7}$$
$$\sqrt{8}$$

y can not be $$\sqrt{9}$$ as y = 3 then x will also be 3.

Its not an ordered pair.

So total values will be 8.

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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 13:39
(x-y)^2+2y^2=18
We are to find integer values of x and y that satisfy the above equation.
(x-y)>4 will produce LHS>RHS
So our combination integer values of x and y which can satisfy this equation must be such that (x-y)<=4.

When y=-1, (x+1)^2+2=18
x^2+2x-15=0
(x-3)(x+5)=0
Hence x=-5 or x=3
So we have (x,y) pairs of (-5,-1) and (3,-1)

When y=1, (x-1)^2+2=18
x^2-2x-15=0
(x+5)(x-3)=0
Hence x=-5 or x=3
We have (x,y) integer pairs of (-5,1) and (3,1).

Every other integral combinations of (x,y) does not result in a LHS which equals 18 (RHS).

Therefore there are 4 ordered pairs (x,y) that are solutions of the equation.

Posted from my mobile device
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 15:44
Here (x−y)^2+2^2=18

So we can try this with number picking method too. Both elements are squared , so they have to be positive.

Now lets say y =0 then after putting this value in above equation we dont get the integer value of x.

Y= 1 , then x= 5,-3 so , here got two pair ( 5,1) (-3,1)
y= 2, then x cant be an integer
y= 3, then then x can be 3 so here the pair is (3,3) now we can have (-3,-3) too

So there are 4 pairs . Ans - B
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 16:13
Start populating with y=0 ---> x integer
y=1 ---> x not integer
y=2---->x not integer
y=3------> x integer.
therefore 1,3 satisfies also -1,-3 will satisfy
total 4 solutions
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For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 17:34
solving for the given eqn ; we get
x^2+3y^2-2xy=18
test with y=1 we get x=5,-3 then at x=5 y=3,7
total pairs ; IMO B ; 4

(x−y)2+2y2=18(x−y)2+2y2=18

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

A. 2
B. 4
C. 6
D. 8
E. 10
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For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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Updated on: 16 Jul 2019, 18:51
1
$$(x−y)^2 + 2y^2 = 18$$ can be rewritten as $$(x−y)^2 = 2*(3 - y)*(3+y)$$

$$(x−y)^2 \geq{0}$$, then $$2*(3 - y)*(3+y) \geq{0}$$, which limits us to $$-3 \leq{y} \leq{3}$$. Given x and y have to be both integers, y can be of only 7 possible values: $$-3, -2, -1, 0, 1, 2, 3$$

$$y=0,$$ $$y=-2$$ and $$y=2$$ all lead to non-integer x. Eliminate them.

There are six integer pairs $$(x , y)$$ that are solutions of the equation as shown below:
$$y=-3,$$ then $$(x+3)^2 =0$$ and $$x=-3$$ --> $$(-3,-3)$$ is a solution.
$$y=3,$$ then $$(x-3)^2 =0$$ and $$x=3$$ --> $$(3,3)$$ is a solution.
$$y=-1,$$ then $$(x+1)^2 =16$$ and $$x=-5$$ or $$3$$ --> $$(-5,-1)$$ and $$(3,-1)$$ are both solution.
$$y=1,$$ then $$(x-1)^2 =16$$ and $$x=5$$ or $$-3$$ --> $$(5,1)$$ and $$(-3,1)$$ are both solution.

Originally posted by chondro48 on 16 Jul 2019, 18:45.
Last edited by chondro48 on 16 Jul 2019, 18:51, edited 2 times in total.
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 18:46
(x - y)^2 + 2 y^2 = 18
x^2 - 2 x y + 3 y^2 - 18 = 0

plug in numbers
x = -5, y = -1
x = -3, y = -3
x = -3, y = 1
x = 3, y = -1
x = 3, y = 3
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 20:02
1
(x−y)^2+2y^2=18

We can start putting values of y:
y=0; (x)^2=18, does not give x as interger
y=1; (x−1)^2+2=18 --> (x−1)^2=16 -> (x−1)=+/-4 -> x=5 or-3
y=-1; (x+1)=+/-4 -> x=3 or -5
y=3; (x−3)=0 -> x=3
y=-3; (x+3)=0 -> x=-3

Ordered pairs of (x,y)
5,1
-3,1
3,-1
-5,1
3,3
-3,3

C is correct.
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 20:12
Given equation is (x−y)^2+2y^2=18 , and x & y are integers

Instead of trying to solve this equation , we can do trail and error method

lets say (x−y)^2+2y^2=18

18 - could be written as 16+2 , 2*9,6*3 , 18*1

But Only 16+2 and 2*9 can will us solutions where x and y are integers

==> case 1 : where x = y = 3 , (3−3)^2+2*3^2=18
Case 2 : where x = y = -3 , (-3+3)^2+2(-3)^2=18
Case 3 : where x = 5 , y=1 , (5-1)^2+2*1^2=18
Case 4 : where x = 3 , y= -1 , (3+1)^2+2(-1)^2=18

Total 4 cases ( 3,3) ,(-3,-3) , (5,1) ,(4,-1)--Option B
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For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 21:15
1
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.

For how many ordered pairs (x , y) that are solutions of the equation   [#permalink] 16 Jul 2019, 21:15

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