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Friday, January 9, 2015

2015/002) show that if 3x^2+x=4y^2+y then each of x-y , 3x+3y+1 and 4x+4y+1 is a perfect square

we have

3x^2+x = 4y^2 + y
or 3x^2- 3y^2 + x - y = y^2
or 3(x+y)(x-y) + (x-y) = y^2
or (3x+3y+1)(x-y) = y^2\cdots\, 1
also
4x^2-4y^2 + x-y=x^2
or 4(x-y)(x+y) + (x-y) = x^2
or (x-y)(4x+4y+1) = x^2\cdots\,2
multiply (1) and (2) to get

(x-y)^2(3x+3y+1)(4x+4y+1)=x^2y^2
or(3x+3y+1)(4x+4y+1)=(\dfrac{xy}{x-y})^2

so (3x+3y+1)(4x+4y+1) is a perfect square

as 4(3x+3y+1) - 3(4x+4y+1) = 1 so (3x+3y+1) and (4x+4y+1) both are perfect squares and the from (1) or (2) (x-y) is a perfect square

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