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Need to find $n,r$ (if any) for $121^r-2n^2=1$ where $n,r$ are natural numbers.

Observed that $n$ is odd then $n=2m+1$ (say).

But on replacement of $n$ by $2m+1$, increases the complexity of the problem.

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    $11^r=2n^2+1$ Hence $2n^2\cong10\mod11$, $n^2\cong5\mod11$, $n\cong4$ or $7\mod11$ thereby eliminating about 82% of the natural numbers. Will this help?2012-07-13

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All solutions to $x^2-2y^2=1$ are given implicitly by $ x_k+y_k\sqrt{2} = (3+2\sqrt{2})^k $ To solve your equation would require $x_k=11^r$ for some $k,r$.

But looking at solutions modulo 11 and 3 we find a cycle of 12 possibilities mod 11 $(x_{13},y_{13})\equiv (3,2)\pmod{11}$ and a cycle of four possibilities mod 3, $(x_5,y_5)\equiv (0,2) \pmod{3}$. Considering divisibility of $x$ we find $11\mid x_k$ only when $k\equiv 3 \pmod{6}$, and $3\mid x_k$ whenever $k$ is odd. Thus $x_k$ cannot be a power of $11$ and your equation has no solutions.

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    Just a comment on notation: the variable $n$ is already used for something else in the OP's question. You might want to change it to avoid any confusion.2012-07-13