Show that the modified Pell equation $x^2 - 7y^2 = -1$ has no solutions in integers $x,y$.

Question

Show that the modified Pell equation $x^2 - 7y^2 = -1$ has no solutions in integers $x,y$. (Hint: reduce the equation modulo a suitably chosen prime.)

I think that we can use the Diophantine equation for this, but I don't know where to start. I am new to this material in Number Theory.

Answer 1

Add $8y^2+4$ to both sides of the equation getting $x^2+y^2+4 = 8y^2+3$ Now read this equation modulo 4, we get $x^2+y^2\equiv3\pmod 4$. As any square leaves a remainder of 0 or 1 mod 4, adding two of them (the LHS) we cannot get 3 (the RHS).

Answer 2

Try lifting all the solutions modulo $7$.

Answer 3

$$ \sqrt { 7} = 2 + \frac{ \sqrt {7} - 2 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {7} - 2 } = \frac{ \sqrt {7} + 2 }{3 } = 1 + \frac{ \sqrt {7} - 1 }{3 } $$ $$ \frac{ 3 }{ \sqrt {7} - 1 } = \frac{ \sqrt {7} + 1 }{2 } = 1 + \frac{ \sqrt {7} - 1 }{2 } $$ $$ \frac{ 2 }{ \sqrt {7} - 1 } = \frac{ \sqrt {7} + 1 }{3 } = 1 + \frac{ \sqrt {7} - 2 }{3 } $$ $$ \frac{ 3 }{ \sqrt {7} - 2 } = \frac{ \sqrt {7} + 2 }{1 } = 4 + \frac{ \sqrt {7} - 2 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccc} & & 2 & & 1 & & 1 & & 1 & & 4 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 2 }{ 1 } & & \frac{ 3 }{ 1 } & & \frac{ 5 }{ 2 } & & \frac{ 8 }{ 3 } \\ \\ & 1 & & -3 & & 2 & & -3 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 7 \cdot 0^2 = 1 & \mbox{digit} & 2 \\ \frac{ 2 }{ 1 } & 2^2 - 7 \cdot 1^2 = -3 & \mbox{digit} & 1 \\ \frac{ 3 }{ 1 } & 3^2 - 7 \cdot 1^2 = 2 & \mbox{digit} & 1 \\ \frac{ 5 }{ 2 } & 5^2 - 7 \cdot 2^2 = -3 & \mbox{digit} & 1 \\ \frac{ 8 }{ 3 } & 8^2 - 7 \cdot 3^2 = 1 & \mbox{digit} & 4 \\ \end{array} $$

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Here's a good one, not amazingly hard. Show that the modified Pell equation $x^2 - 221y^2 = -1$ has no solutions in integers $x,y$.

$$ \sqrt { 221} = 14 + \frac{ \sqrt {221} - 14 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {221} - 14 } = \frac{ \sqrt {221} + 14 }{25 } = 1 + \frac{ \sqrt {221} - 11 }{25 } $$ $$ \frac{ 25 }{ \sqrt {221} - 11 } = \frac{ \sqrt {221} + 11 }{4 } = 6 + \frac{ \sqrt {221} - 13 }{4 } $$ $$ \frac{ 4 }{ \sqrt {221} - 13 } = \frac{ \sqrt {221} + 13 }{13 } = 2 + \frac{ \sqrt {221} - 13 }{13 } $$ $$ \frac{ 13 }{ \sqrt {221} - 13 } = \frac{ \sqrt {221} + 13 }{4 } = 6 + \frac{ \sqrt {221} - 11 }{4 } $$ $$ \frac{ 4 }{ \sqrt {221} - 11 } = \frac{ \sqrt {221} + 11 }{25 } = 1 + \frac{ \sqrt {221} - 14 }{25 } $$ $$ \frac{ 25 }{ \sqrt {221} - 14 } = \frac{ \sqrt {221} + 14 }{1 } = 28 + \frac{ \sqrt {221} - 14 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccccccc} & & 14 & & 1 & & 6 & & 2 & & 6 & & 1 & & 28 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 14 }{ 1 } & & \frac{ 15 }{ 1 } & & \frac{ 104 }{ 7 } & & \frac{ 223 }{ 15 } & & \frac{ 1442 }{ 97 } & & \frac{ 1665 }{ 112 } \\ \\ & 1 & & -25 & & 4 & & -13 & & 4 & & -25 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 221 \cdot 0^2 = 1 & \mbox{digit} & 14 \\ \frac{ 14 }{ 1 } & 14^2 - 221 \cdot 1^2 = -25 & \mbox{digit} & 1 \\ \frac{ 15 }{ 1 } & 15^2 - 221 \cdot 1^2 = 4 & \mbox{digit} & 6 \\ \frac{ 104 }{ 7 } & 104^2 - 221 \cdot 7^2 = -13 & \mbox{digit} & 2 \\ \frac{ 223 }{ 15 } & 223^2 - 221 \cdot 15^2 = 4 & \mbox{digit} & 6 \\ \frac{ 1442 }{ 97 } & 1442^2 - 221 \cdot 97^2 = -25 & \mbox{digit} & 1 \\ \frac{ 1665 }{ 112 } & 1665^2 - 221 \cdot 112^2 = 1 & \mbox{digit} & 28 \\ \end{array} $$

Answer 4

Take both sides modulo 7. We have $x^2=-1$ modulo 7 which has no solution as 7 is a prime of type 3 mod 4.