Theorem: Solutions to Diophantine Equation 16x^2+32x+20 = y^2+y

The indeterminate Diophantine equation:
:$16x^2 + 32x + 20 = y^2 + y$
has exactly $4$ solutions:
:$\tuple {0, 4}, \tuple {-2, 4}, \tuple {0, -5}, \tuple {-2, -5}$

Proof:
{{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}}