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 Feb20 awarded Yearling May20 awarded Caucus Mar28 revised Curious $\sum _{n=1}^{\infty} \frac{1}{n^2 - x^2}$ identity \bigg paranthesis :) Mar28 suggested approved edit on Curious $\sum _{n=1}^{\infty} \frac{1}{n^2 - x^2}$ identity Mar27 awarded Commentator Mar27 comment What's wrong in this equation? @StefanHansen, your answer is completely correct. Inceptio we know nothing about $x$. It is not mentioned that $x$ is an integer, or anything. Stefan makes a very valid point about that, by demonstrating the case of 2.5. This is the problem in the proof, we cannot say '$\text{x times}$'. Mar27 answered What's wrong in this equation? Mar25 comment Mathematical way of determining whether a number is an integer Yes, floor is a purely mathematical function. Mar24 revised $x$ is rational, $\frac{x}{2}$ is rational, and $3x-1$ is rational are equivalent Improved definition Mar24 suggested approved edit on $x$ is rational, $\frac{x}{2}$ is rational, and $3x-1$ is rational are equivalent Mar24 comment $x$ is rational, $\frac{x}{2}$ is rational, and $3x-1$ is rational are equivalent What is your definition of rational? If you can answer that, the result is immediate. Mar24 accepted Twice a triangle is triangle Mar23 comment Twice a triangle is triangle @AndréNicolas, the Pell's equation I have is $x^2 - 2y^2 = -1$, and not $x^2 - 2y^2 = 1$. I know how to do the latter, squaring and re-arranging gives the recurrence relations. Its the case of $-1$ which I can't deal with. Squaring gives extraneous incorrect solutions, and cubing gives unhelpful result. Mar22 awarded Editor Mar22 comment Twice a triangle is triangle @Alvaro No, not closed form, in terms of the previous solutions. Mar22 revised Twice a triangle is triangle corrected link Mar22 comment Twice a triangle is triangle I already read this and similar questions, but they don't tell how to solve for the recurrence. Mar22 asked Twice a triangle is triangle Mar22 answered inscribed angles on circle Mar18 comment Establishing formula from recurrence With the help of your excellent response, I managed to write a general form of $f(n)$, ie $f(n) = (y + \frac{b}{a-1}) \cdot a^{n-x} - \frac{b}{a-1}$ given $f(x) = y$ (base condition) and $f(n) = a\cdot f(n-1) + b$. In the case of the given problem, $f(n) = (1 + \frac{3}{-4-1}) \cdot (-4)^{n-1} - \frac{3}{-4-1}$ which simplifies to $f(n) = \frac{2}{5} \cdot (-4)^{n-1} + \frac{3}{5}$. Thanks a lot for your help.