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I know how to solve linear diophantine equations, but I was wondering if someone can give me a step by step to solve something like $2x^2 + 2x - 5y = -1$? I cannot find a lot of resources on this particular form.

I know the solutions are

$${ y = 10 k^2 - 14 k + 5, x = 3 - 5 k}$$

$${ y = 10 k^2 - 6 k + 1, x = 1 - 5 k}$$

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$$ 2 x^2 + 2 x + 1 \equiv 0 \pmod 5 $$ $$ 4 x^2 + 4 x + 2 \equiv 0 \pmod 5 $$ $$ (4 x^2 + 4 x + 1) +1 \equiv 0 \pmod 5 $$ $$ (4 x^2 + 4 x + 1) \equiv -1 \pmod 5 $$ $$ (2 x + 1)^2 \equiv -1 \pmod 5 $$ $$ 2 x + 1 \equiv 2,3 \pmod 5 $$ $$ 2 x \equiv 1,2 \pmod 5 $$ $$ 2 x \equiv 6,2 \pmod 5 $$ $$ x \equiv 3,1 \pmod 5 $$

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  • $\begingroup$ Thanks alot, quick question though. The step where you get rid of the square is through Modular multiplicative inverse? $\endgroup$ – Ryan Topps Sep 28 '18 at 3:33
  • $\begingroup$ @RyanTopps I wouldn't put it that way. If we are considering things $\pmod p$ where $p$ is an odd prime, square roots of a number (if there are any at all) come in $\pm$ pairs. As $-1 \equiv 4 \equiv 9 \pmod 5,$ we find the square roots as $2,3 \pmod 5.$ Note that $2+3 \equiv 0 \pmod 5$ $\endgroup$ – Will Jagy Sep 28 '18 at 3:35
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Methods for solving equations in integers:

https://youtu.be/MCCKGXYMSj8 (with an error)

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  • $\begingroup$ It would be helpful if you summarized the material in the link. This is useful in case the link were to go dead at some point. It's especially important if the linked video contains an error. $\endgroup$ – Andrew Uzzell Sep 28 '18 at 15:03

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