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