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For example; $$y^2 = 30x + 1$$ For which one answer is; $$11^2 = 30(4) + 1 = 121$$

WA kindly gave me 4 answers; $$x = 2 (15 n^2 - 29 n + 14), y = 29 - 30 n, n \in Z$$ $$x = 2 (15 n^2 - 19 n + 6), y = 19 - 30 n, n \in Z$$ $$x = 2 (15 n^2 - 11 n + 2), y = 11 - 30 n, n \in Z$$ $$x = 2 (15 n^2 - n), y = 1 - 30 n, n \in Z$$

Please help me learn how to get from my example to those 4 answers.

I'm more interested in a general solution in terms of $m$ and $b$. $30$ and $1$ are just random, convenient values for example. Though, I can work backward from example values to the general solution.

I recognize the two coefficients of $n$ being $1$ and $29$ are trivially $b$ and $m-b$, respectively.

I'm particularly interested in whether there's an algebraic calculation that yields the coefficients of $11$ and $19$, or whether it's a search function. I.e., what if $m$ were very large?

CAB
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  • Do you want to find every solution to $y^2=30x+1$? Or do you want a method you can use in order to find one solution? – Dr. Mathva Oct 21 '20 at 14:54
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    If you were to consider the equation modulo $30$ you would have $y^2\equiv 1\pmod{30}$ which heavily limits things. – JMoravitz Oct 21 '20 at 14:55
  • @Dr.Mathva - I added to my post. Hope it clarifies my ask. – CAB Oct 22 '20 at 15:59
  • How does this differ from the question you asked a year ago, https://math.stackexchange.com/questions/3276284/how-to-find-integer-solutions-of-x2-by-c ? – Barry Cipra Oct 22 '20 at 16:20

2 Answers2

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Notice that your equation yields $$x=\frac{y^2-1}{30}\implies y^2\equiv1\bmod 30\iff (y+1)(y-1)\equiv 0\bmod 30$$ It is simple to infer that $y\bmod 30\in\{\pm1, \pm11\} $. If $y=30k+1\;, k\in\mathbb Z $ then $$x=30k^2-2k$$ Therefore, $(x,y)=(30k^2-2k, 30k+1)$ for every $k\in\mathbb Z$ are all the solutions when $y\equiv 1\bmod 30$. Can you work out the other three cases?

Dr. Mathva
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$y^2 = mx +b$ so $x=\frac {y^2-b}m$ needs to be an integer.

I don't know of any general way to do it but you could solve $y^2 = b\pmod m$ (which may or may not have tricks; but brute force we can test $m$ values). $\alpha$ is a such a solution then $\alpha + mk$ will be solutions.

For example for $y^2 = 6m + 7$ we must have $y^2 \equiv 7\equiv 1\pmod 6$ and and so $y^2-1\equiv (y-1)(y+1)\equiv 0\pmod 6$ so $y\equiv 1, -1$ and $y-1\equiv 2,3$ while $y+1\equiv 3,2$ but the latter are impossible so $y=6k\pm 1$ will be a solution.

Less neat example $y^2 = 7m + 12$ we must have $y^2 \equiv 12\equiv 5\pmod 7$. Brute force $0^2,(\pm1)^2,(\pm 2)^2,(\pm 3)^2\equiv 0,1,4,2$ so that has no solutions. But for $y^2 = 7m + 11$ we'd have $y=7m \pm 2$ always a solution.

Verification $(7m\pm 2)^2 = 49m^2 \pm 28m + 4= 7(7m^2 \pm 4m-1) + 11$. So $x=7m^2\pm 4m-1; y=7m\pm 2$ is a solution to $y^2 = 7x +11$

And for $k=0....6$ $(7m+k)^2 = 49m^2 +14mk + k^2 = 7(7m^2+2mk) +k^2$ but for $k^2\ne 7j + 12$ for any $j$ so there are no integer solutions to $y^2=7x + 12$.

fleablood
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