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

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Theorem

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

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


$17$ 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:

$\tuple {-2, -5}$

$\blacksquare$


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