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# PS - equation.

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Manager
Joined: 09 Sep 2004
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09 Sep 2004, 10:16
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Hi everyone,
Can anybody explain how to solve this problem.

If x and y are integers , then x = ?

(x - y)^2 + 2y^2 = 27.

Thanks.
Manager
Joined: 05 Sep 2004
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09 Sep 2004, 10:51
If x and y are integers, the it follows that:

(i) (x-y)^2 is a positive integer
(ii) 2(y^2) is a positive integer

This means that 2(y^2) has to be less than 27.
=> So, y can only assume one of the three values: 0, +-1, +-2, and +-3. (For example, if y = 4, y^2 is 16, and 2(y^2) is 32, which is more than 27.

Now, if y = 0, 2(0^2) is also 0, and then (x-0)^2 = 27, => x^2 = 27, which is impossible if x is an integer.

If y = +-1, 2(+-1^2) = 2, and then [x-(+-1)]^2 = 25, => x-(+-1) = 5.
Thus, either x-1 = 5, or x+1 = 5, which results into x = 6, or x = 4

If y = +-2, 2(+-2^2) = 8, and then [x-(+-2)]^2 = 19, which is not a perfect square (i.e., solution of x will not be an integer).

If y = +-3, 2(+-3^2) = 18, and then [x-(+-3)]^2 = 9, => x-(+-3) = 3.
Thus, either x-3 = 3, or x+3 = 3, which results into x = 6, or x = 0.

Summarizing, the following are the possible solutions to the given equation:

(x,y) = (6,1), (4, -1), (6,3), (0, -3)

So, x can be any of 0, 4, and 6.

All of the above values satisfy the given equation.
Manager
Joined: 16 Jul 2003
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10 Sep 2004, 09:48
intr3pid wrote:
If x and y are integers, the it follows that:

(i) (x-y)^2 is a positive integer
(ii) 2(y^2) is a positive integer

This means that 2(y^2) has to be less than 27.
=> So, y can only assume one of the three values: 0, +-1, +-2, and +-3. (For example, if y = 4, y^2 is 16, and 2(y^2) is 32, which is more than 27.

Now, if y = 0, 2(0^2) is also 0, and then (x-0)^2 = 27, => x^2 = 27, which is impossible if x is an integer.

If y = +-1, 2(+-1^2) = 2, and then [x-(+-1)]^2 = 25, => x-(+-1) = 5.
Thus, either x-1 = 5, or x+1 = 5, which results into x = 6, or x = 4

If y = +-2, 2(+-2^2) = 8, and then [x-(+-2)]^2 = 19, which is not a perfect square (i.e., solution of x will not be an integer).

If y = +-3, 2(+-3^2) = 18, and then [x-(+-3)]^2 = 9, => x-(+-3) = 3.
Thus, either x-3 = 3, or x+3 = 3, which results into x = 6, or x = 0.

Summarizing, the following are the possible solutions to the given equation:

(x,y) = (6,1), (4, -1), (6,3), (0, -3)

So, x can be any of 0, 4, and 6.

All of the above values satisfy the given equation.

Thanks for the explanation. One stupid question - "This means that 2(y^2) has to be less than 27" How did you deduce this?
Director
Joined: 20 Jul 2004
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11 Sep 2004, 11:20
stuti wrote:
Thanks for the explanation. One stupid question - "This means that 2(y^2) has to be less than 27" How did you deduce this?

Because, as intrepid said,
(i) (x-y)^2 is a positive integer
(ii) 2(y^2) is a positive integer

When you add two positive values they should each be lesses than their sum. (for positive a and b, if a+b=c, then a and b should be less than c)
11 Sep 2004, 11:20
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