{{begin-eqn}} {{eqn | l = 16 x^2 + 32 x + 20        | r = y^2 + y       | c = }} {{eqn | ll= \leadsto       | l = 16 x^2 + 32 x + 16 + 4       | r =       | c = }} {{eqn | l = 16 \paren {x^2 + 2 x + 1} + 4       | r =       | c = }} {{eqn | l = 16 \paren {x + 1}^2 + 4       | r = y^2 + y       | c = }} {{eqn | ll= \leadsto       | l = 64 \paren {x + 1}^2 + 16       | r = 4 y^2 + 4 y       | c = }} {{eqn | ll= \leadsto       | l = \paren {8 x + 8}^2 + 17       | r = 4 y^2 + 4 y + 1       | c = }} {{eqn | ll= \leadsto       | l = \paren {8 x + 8}^2 + 17       | r = \paren {2 y + 1}^2       | c = }} {{eqn | ll= \leadsto       | l = 17       | r = \paren {2 y + 1}^2 - \paren {8 x + 8}^2       | c = }} {{eqn | r = \paren {2 y + 1 - 8 x - 8} \paren {2 y + 1 + 8 x + 8}       | c = }} {{eqn | r = \paren {2 y - 8 x - 7} \paren {2 y + 8 x + 9}       | c = }} {{end-eqn}} $17$ is prime and therefore the solution of only two sets of integer products: {{begin-eqn}} {{eqn | l = 17       | r = 1 \times 17       | c = }} {{eqn | l = 17       | r = -1 \times -17       | c = }} {{end-eqn}} This leaves us with four systems of equations with four solutions: {{begin-eqn}} {{eqn | l = 1       | r = 2 y - 8 x - 7       | c = }} {{eqn | l = 17       | r = 2 y + 8 x + 9       | c = }} {{eqn | ll= \leadsto       | l = 1 + 17       | r = 2y - 8x + 9 + 2y + 8x - 7       | c = }} {{eqn | ll= \leadsto       | l = 18       | r = 4y + 2       | c = }} {{eqn | ll= \leadsto       | l = 4       | r = y       | c = }} {{eqn | l = 1       | r = 2 \paren 4 - 8 x - 7       | c = }} {{eqn | ll= \leadsto       | l = 0       | r = x       | c = }} {{end-eqn}} Hence the solution: :$\tuple {0, 4}$ {{begin-eqn}} {{eqn | l = 17       | r = 2 y - 8 x - 7       | c = }} {{eqn | l = 1       | r = 2 y + 8 x + 9       | c = }} {{eqn | ll= \leadsto       | l = 17 + 1       | r = 2 y - 8 x - 7 + 2 y + 8 x + 9       | c = }} {{eqn | ll= \leadsto       | l = 18       | r = 4 y + 2       | c = }} {{eqn | ll= \leadsto       | l = 4       | r = y       | c = }} {{eqn | l = 1       | r = 2 \tuple 4 + 8 x + 9       | c = }} {{eqn | ll= \leadsto       | l = -2       | r = x       | c = }} {{end-eqn}} Hence the solution: :$\tuple {-2, 4}$ {{begin-eqn}} {{eqn | l = -17       | r = 2 y - 8 x - 7       | c = }} {{eqn | l = -1       | r = 2 y + 8 x + 9       | c = }} {{eqn | ll= \leadsto       | l = -1 - 17       | r = 2 y - 8 x + 9 + 2 y + 8 x - 7       | c = }} {{eqn | ll= \leadsto       | l = -18       | r = 4 y + 2       | c = }} {{eqn | ll= \leadsto       | l = -5       | r = y       | c = }} {{eqn | l = -1       | r = -2 \paren {-5} - 8 x - 7       | c = }} {{eqn | ll= \leadsto       | l = 0       | r = x       | c = }} {{end-eqn}} Hence the solution: :$\tuple {0, -5}$ {{begin-eqn}} {{eqn | l = -1       | r = 2 y - 8 x - 7       | c = }} {{eqn | l = -17       | r = 2 y + 8 x + 9       | c = }} {{eqn | ll= \leadsto       | l = -1 - 17       | r = 2 y - 8 x + 9 + 2 y + 8 x - 7       | c = }} {{eqn | ll= \leadsto       | l = -18       | r = 4 y + 2       | c = }} {{eqn | ll= \leadsto       | l = -5       | r = y       | c = }} {{eqn | l = -1       | r = 2 \paren {-5} - 8 x - 7       | c = }} {{eqn | ll= \leadsto       | l = -2       | r = x       | c = }} {{end-eqn}} Hence the solution: :$\tuple {-2, -5}$ {{qed}}