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

## Theorem

$16x^2 + 32x + 20 = y^2 + y$

has exactly $4$ solutions:

$\tuple {0, 4}, \tuple {-2, 4}, \tuple {0, -5}, \tuple {-2, -5}$

## Proof

 $\displaystyle 16 x^2 + 32 x + 20$ $=$ $\displaystyle y^2 + y$ $\displaystyle \leadsto \ \$ $\displaystyle 16 x^2 + 32 x + 16 + 4$ $=$ $\displaystyle$ $\displaystyle 16 \paren {x^2 + 2 x + 1} + 4$ $=$ $\displaystyle$ $\displaystyle 16 \paren {x + 1}^2 + 4$ $=$ $\displaystyle y^2 + y$ $\displaystyle \leadsto \ \$ $\displaystyle 64 \paren {x + 1}^2 + 16$ $=$ $\displaystyle 4 y^2 + 4 y$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {8 x + 8}^2 + 17$ $=$ $\displaystyle 4 y^2 + 4 y + 1$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {8 x + 8}^2 + 17$ $=$ $\displaystyle \paren {2 y + 1}^2$ $\displaystyle \leadsto \ \$ $\displaystyle 17$ $=$ $\displaystyle \paren {2 y + 1}^2 - \paren {8 x + 8}^2$ $\displaystyle$ $=$ $\displaystyle \paren {2 y + 1 - 8 x - 8} \paren {2 y + 1 + 8 x + 8}$ $\displaystyle$ $=$ $\displaystyle \paren {2 y - 8 x - 7} \paren {2 y + 8 x + 9}$

$17$ is prime and therefore the solution of only two sets of integer products:

 $\displaystyle 17$ $=$ $\displaystyle 1 \times 17$ $\displaystyle 17$ $=$ $\displaystyle -1 \times -17$

This leaves us with four systems of equations with four solutions:

 $\displaystyle 1$ $=$ $\displaystyle 2 y - 8 x - 7$ $\displaystyle 17$ $=$ $\displaystyle 2 y + 8 x + 9$ $\displaystyle \leadsto \ \$ $\displaystyle 1 + 17$ $=$ $\displaystyle 2y - 8x + 9 + 2y + 8x - 7$ $\displaystyle \leadsto \ \$ $\displaystyle 18$ $=$ $\displaystyle 4y + 2$ $\displaystyle \leadsto \ \$ $\displaystyle 4$ $=$ $\displaystyle y$ $\displaystyle 1$ $=$ $\displaystyle 2 \paren 4 - 8 x - 7$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $=$ $\displaystyle x$

Hence the solution:

$\tuple {0, 4}$

 $\displaystyle 17$ $=$ $\displaystyle 2 y - 8 x - 7$ $\displaystyle 1$ $=$ $\displaystyle 2 y + 8 x + 9$ $\displaystyle \leadsto \ \$ $\displaystyle 17 + 1$ $=$ $\displaystyle 2 y - 8 x - 7 + 2 y + 8 x + 9$ $\displaystyle \leadsto \ \$ $\displaystyle 18$ $=$ $\displaystyle 4 y + 2$ $\displaystyle \leadsto \ \$ $\displaystyle 4$ $=$ $\displaystyle y$ $\displaystyle 1$ $=$ $\displaystyle 2 \tuple 4 + 8 x + 9$ $\displaystyle \leadsto \ \$ $\displaystyle -2$ $=$ $\displaystyle x$

Hence the solution:

$\tuple {-2, 4}$

 $\displaystyle -17$ $=$ $\displaystyle 2 y - 8 x - 7$ $\displaystyle -1$ $=$ $\displaystyle 2 y + 8 x + 9$ $\displaystyle \leadsto \ \$ $\displaystyle -1 - 17$ $=$ $\displaystyle 2 y - 8 x + 9 + 2 y + 8 x - 7$ $\displaystyle \leadsto \ \$ $\displaystyle -18$ $=$ $\displaystyle 4 y + 2$ $\displaystyle \leadsto \ \$ $\displaystyle -5$ $=$ $\displaystyle y$ $\displaystyle -1$ $=$ $\displaystyle -2 \paren {-5} - 8 x - 7$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $=$ $\displaystyle x$

Hence the solution:

$\tuple {0, -5}$

 $\displaystyle -1$ $=$ $\displaystyle 2 y - 8 x - 7$ $\displaystyle -17$ $=$ $\displaystyle 2 y + 8 x + 9$ $\displaystyle \leadsto \ \$ $\displaystyle -1 - 17$ $=$ $\displaystyle 2 y - 8 x + 9 + 2 y + 8 x - 7$ $\displaystyle \leadsto \ \$ $\displaystyle -18$ $=$ $\displaystyle 4 y + 2$ $\displaystyle \leadsto \ \$ $\displaystyle -5$ $=$ $\displaystyle y$ $\displaystyle -1$ $=$ $\displaystyle 2 \paren {-5} - 8 x - 7$ $\displaystyle \leadsto \ \$ $\displaystyle -2$ $=$ $\displaystyle x$

Hence the solution:

$\tuple {-2, -5}$

$\blacksquare$