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the number of ordered pairs of positive integers $x,$y such that $x^2 +3y$ and $y^2 +3x$
are both perfect squares
my solution::
-3(x-y)=k^2-l^2}$\\\\%20$\bf{(x-y).(x+y-3)=(k+l).(k-l)}$\\\\%20$\bullet\;\;%20\bf{(x-y)=k+l\;\;,(x+y-3)=k-l}$\\\\%20$\bullet\;\;%20\bf{(x-y)=k-l\;\;,(x+y-3)=k+l}$\\\\%20So%20$\bf{x=\frac{2k+3}{2}\notin%20\mathbb{Z^{+}}}$\\\\%20and%20$\bf{y=\frac{-2l+3}{2}\notin%20\mathbb{Z^{+}}}$\\\\)
no possibilities.
but there is also more possibilities
like $(x-y).(x+y-3) = 1 \times (k^2-l^2) = (k^2-l^2) \times 1$
My Question is that is any pairs for which $x^2+3y$ and $3x^2+y$ are perfect square
Thanks
are both perfect squares
my solution::
no possibilities.
but there is also more possibilities
like $(x-y).(x+y-3) = 1 \times (k^2-l^2) = (k^2-l^2) \times 1$
My Question is that is any pairs for which $x^2+3y$ and $3x^2+y$ are perfect square
Thanks
