- #1
juantheron
- 247
- 1
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::
http://latex.codecogs.com/gif.latex?\hspace{-16}$Let%20$\bf{x^2+3y=k^2}$%20and%20$\bf{y^2+3x=l^2}$\\%20Where%20$\bf{x,y,k,l\in%20\mathbb{Z^{+}}}$\\%20$\bf{(x^2-y^2)-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::
http://latex.codecogs.com/gif.latex?\hspace{-16}$Let%20$\bf{x^2+3y=k^2}$%20and%20$\bf{y^2+3x=l^2}$\\%20Where%20$\bf{x,y,k,l\in%20\mathbb{Z^{+}}}$\\%20$\bf{(x^2-y^2)-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