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

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


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

\(\ds 17\) \(=\) \(\ds 1 \times 17\)
\(\ds 17\) \(=\) \(\ds -1 \times -17\)


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

\(\ds 1\) \(=\) \(\ds 2 y - 8 x - 7\)
\(\ds 17\) \(=\) \(\ds 2 y + 8 x + 9\)
\(\ds \leadsto \ \ \) \(\ds 1 + 17\) \(=\) \(\ds 2y - 8x + 9 + 2y + 8x - 7\)
\(\ds \leadsto \ \ \) \(\ds 18\) \(=\) \(\ds 4y + 2\)
\(\ds \leadsto \ \ \) \(\ds 4\) \(=\) \(\ds y\)
\(\ds 1\) \(=\) \(\ds 2 \paren 4 - 8 x - 7\)
\(\ds \leadsto \ \ \) \(\ds 0\) \(=\) \(\ds x\)

Hence the solution:

$\tuple {0, 4}$


\(\ds 17\) \(=\) \(\ds 2 y - 8 x - 7\)
\(\ds 1\) \(=\) \(\ds 2 y + 8 x + 9\)
\(\ds \leadsto \ \ \) \(\ds 17 + 1\) \(=\) \(\ds 2 y - 8 x - 7 + 2 y + 8 x + 9\)
\(\ds \leadsto \ \ \) \(\ds 18\) \(=\) \(\ds 4 y + 2\)
\(\ds \leadsto \ \ \) \(\ds 4\) \(=\) \(\ds y\)
\(\ds 1\) \(=\) \(\ds 2 \tuple 4 + 8 x + 9\)
\(\ds \leadsto \ \ \) \(\ds -2\) \(=\) \(\ds x\)

Hence the solution:

$\tuple {-2, 4}$


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

Hence the solution:

$\tuple {0, -5}$


\(\ds -1\) \(=\) \(\ds 2 y - 8 x - 7\)
\(\ds -17\) \(=\) \(\ds 2 y + 8 x + 9\)
\(\ds \leadsto \ \ \) \(\ds -1 - 17\) \(=\) \(\ds 2 y - 8 x + 9 + 2 y + 8 x - 7\)
\(\ds \leadsto \ \ \) \(\ds -18\) \(=\) \(\ds 4 y + 2\)
\(\ds \leadsto \ \ \) \(\ds -5\) \(=\) \(\ds y\)
\(\ds -1\) \(=\) \(\ds 2 \paren {-5} - 8 x - 7\)
\(\ds \leadsto \ \ \) \(\ds -2\) \(=\) \(\ds x\)

Hence the solution:

$\tuple {-2, -5}$

$\blacksquare$


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