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Question Number 73033 by mathmax by abdo last updated on 05/Nov/19
solve inside N^2     x(x+1)=4y(y+1)
$${solve}\:{inside}\:{N}^{\mathrm{2}} \:\:\:\:{x}\left({x}+\mathrm{1}\right)=\mathrm{4}{y}\left({y}+\mathrm{1}\right) \\ $$
Answered by mind is power last updated on 05/Nov/19
⇔4x(x+1)=16y(y+1)  ⇒x≥y≥0  ⇔(2x+1)^2 −1=(4y+2)^2 −4  ⇒(4y+2)^2 −(2x+1)^2 =3  ⇒(4y−2x+1)(4y+2x+3)=3  ⇒ { (( 4y−2x+1=1)),((4y+2x+3=3)) :}  ⇒x=y=0
$$\Leftrightarrow\mathrm{4x}\left(\mathrm{x}+\mathrm{1}\right)=\mathrm{16y}\left(\mathrm{y}+\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{x}\geqslant\mathrm{y}\geqslant\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}=\left(\mathrm{4y}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4} \\ $$$$\Rightarrow\left(\mathrm{4y}+\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{3} \\ $$$$\Rightarrow\left(\mathrm{4y}−\mathrm{2x}+\mathrm{1}\right)\left(\mathrm{4y}+\mathrm{2x}+\mathrm{3}\right)=\mathrm{3} \\ $$$$\Rightarrow\begin{cases}{\:\mathrm{4y}−\mathrm{2x}+\mathrm{1}=\mathrm{1}}\\{\mathrm{4y}+\mathrm{2x}+\mathrm{3}=\mathrm{3}}\end{cases} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{y}=\mathrm{0} \\ $$$$ \\ $$

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