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Find the number of ordered pairs (x; y), with x and y both integers, thatsatisfy the...

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christiano-cr7 | Salutatorian

Posted September 21, 2013 at 5:42 AM via web

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Find the number of ordered pairs (x; y), with x and y both integers, that
satisfy the equation: x^2-4y^2 = -3

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 21, 2013 at 5:55 AM (Answer #1)

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The ordered pairs (x, y) where both x and y are integers have to be determined such that x^2 - 4y^2 = -3

x^2 - 4y^2 = -3

=> (x - 2y)(x + 2y) = -3

This is true only if x = 1 and y = 1

The required ordered pair is (1, 1)

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aruv | High School Teacher | Valedictorian

Posted September 21, 2013 at 6:50 AM (Answer #2)

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`x^2-4y^2=-3`

`(i) x=1 ,y=1`

`(1)^2-4(1)^2=-3`

`(ii) x=-1 ,y=1`

`(-1)^2-4(1)^2=-3`

`(iii) x=-1 ,y=-1`

`(-1)^2-4(-1)^2=-3`

`(iv) x=1 ,y=-1`

`(1)^2-4(-1)^2=-3`

Thus  `x^2-4y^2=-3`  has four solutions (1,1),(-1,1),(-1,-1), and (1,-1).

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aruv | High School Teacher | Valedictorian

Posted September 21, 2013 at 8:08 AM (Reply #1)

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how to solve this problem ?

`x^2+3=4y^2`

`x^2+3=(+-sqrt(4y^2))^2`

`x^2+3=(+-2y)^2`

since x and y are integers only

This implies

`x^2=(+-2y)^2-3>=1`

`=> y=+-1`

`=> x^2=1`

`x=+-1`

`Thus`

`(1,1),(1,-1),(-1,-1),(-1,1)`  are possible solution.

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