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Algebra Difficulty 6.1 National olympiad Prove it Greece

Determine all the pairs of real numbers (x,y)(x, y), which are solutions of the system:
{(x2+y2)2−xy(x+y)2=19∣x−y∣=1 \begin{cases} (x^2 + y^2)^2 - xy(x+y)^2 = 19 \\ |x-y| = 1 \end{cases}

Solution

Performing some manipulation into the first equation we find:
{(x2+y2)2−xy(x+y)2=19∣x−y∣=1}⇔{x4+2x2y2+y4−x3y−2x2y2−xy3=19∣x−y∣=1}⇔{x4+y4−x3y−xy3=19∣x−y∣=1}⇔{(x−y)(x3−y3)=19∣x−y∣=1}⇔{(x−y)2(x2+xy+y2)=19∣x−y∣=1}⇔{x2+xy+y2=19∣x−y∣=1}. \left\{ \begin{array}{l} (x^2 + y^2)^2 - xy(x+y)^2 = 19 \\ |x-y| = 1 \end{array} \right\} \Leftrightarrow \left\{ \begin{array}{l} x^4 + 2x^2y^2 + y^4 - x^3y - 2x^2y^2 - xy^3 = 19 \\ |x-y| = 1 \end{array} \right\} \Leftrightarrow \\ \left\{ \begin{array}{l} x^4 + y^4 - x^3y - xy^3 = 19 \\ |x-y| = 1 \end{array} \right\} \Leftrightarrow \left\{ \begin{array}{l} (x-y)(x^3-y^3) = 19 \\ |x-y| = 1 \end{array} \right\} \Leftrightarrow \left\{ \begin{array}{l} (x-y)^2(x^2+xy+y^2) = 19 \\ |x-y| = 1 \end{array} \right\} \\ \Leftrightarrow \left\{ \begin{array}{l} x^2 + xy + y^2 = 19 \\ |x-y| = 1 \end{array} \right\}.
Thus we have the following cases:
∙{x−y=1x2+xy+y2=19}⇔{y=x−1x2+xy+y2=19}⇔{y=x−1x2+x2−x+(x−1)2=19}{y=x−13x2−3x−18=0}⇔{y=x−1x=3 or x=−2}⇔(x,y)=(3,2) or (x,y)=(−2,−3). \bullet \quad \left\{ \begin{array}{l} x - y = 1 \\ x^2 + xy + y^2 = 19 \end{array} \right\} \Leftrightarrow \left\{ \begin{array}{l} y = x - 1 \\ x^2 + xy + y^2 = 19 \end{array} \right\} \Leftrightarrow \left\{ \begin{array}{l} y = x - 1 \\ x^2 + x^2 - x + (x-1)^2 = 19 \end{array} \right\} \\ \left\{ \begin{array}{l} y = x - 1 \\ 3x^2 - 3x - 18 = 0 \end{array} \right\} \Leftrightarrow \left\{ \begin{array}{l} y = x - 1 \\ x = 3 \text{ or } x = -2 \end{array} \right\} \Leftrightarrow (x,y) = (3,2) \text{ or } (x,y) = (-2,-3).
∙{x−y=−1x2+xy+y2=19}⇔{y=x+1x2+xy+y2=19}⇔{y=x+1x2+x2+x+(x+1)2=19}{y=x+13x2+3x−18=0}⇔{y=x+1x=−3 or x=2}⇔(x,y)=(−3,−2) or (x,y)=(2,3). \bullet \quad \left\{ \begin{array}{l} x - y = -1 \\ x^2 + xy + y^2 = 19 \end{array} \right\} \Leftrightarrow \left\{ \begin{array}{l} y = x + 1 \\ x^2 + xy + y^2 = 19 \end{array} \right\} \Leftrightarrow \left\{ \begin{array}{l} y = x + 1 \\ x^2 + x^2 + x + (x+1)^2 = 19 \end{array} \right\} \\ \left\{ \begin{array}{l} y = x + 1 \\ 3x^2 + 3x - 18 = 0 \end{array} \right\} \Leftrightarrow \left\{ \begin{array}{l} y = x + 1 \\ x = -3 \text{ or } x = 2 \end{array} \right\} \Leftrightarrow (x,y) = (-3,-2) \text{ or } (x,y) = (2,3).

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