# Ivan

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 May7 comment Compute the square root of $2$ in $R=\mathbb{Z}/n\mathbb{Z}$ where $n=2^s+1$ and $s=2^k$The reason I am so curious is because @Justin's formula and the entire discussion here doesn't care that $n$ is a Fermat number, but only that $n$ is divisible by 4. May7 comment Compute the square root of $2$ in $R=\mathbb{Z}/n\mathbb{Z}$ where $n=2^s+1$ and $s=2^k$@Jyrki : Is there a danger that $ux \equiv -x$, i.e. that there might be repeats among $x,-x,ux,-ux$? May7 comment Compute the square root of $2$ in $R=\mathbb{Z}/n\mathbb{Z}$ where $n=2^s+1$ and $s=2^k$Sorry, what I meant was that there are for sure 2 distinct solutions given by $x_1=2^{2^{k-2}}(2^{2^{k-1}}-1)$ and $x_2=2^{2^k}+1-x_1$. I was wondering if there are other roots besides these 2, I can't seem to find an argument against such a possibility. May7 comment Compute the square root of $2$ in $R=\mathbb{Z}/n\mathbb{Z}$ where $n=2^s+1$ and $s=2^k$EDIT: Sorry for the double comment. May7 comment Compute the square root of $2$ in $R=\mathbb{Z}/n\mathbb{Z}$ where $n=2^s+1$ and $s=2^k$Is there another square root besides the $\pm$ of the solution above? May6 accepted Non-uniqueness of solutions of the heat equation May6 accepted Generalized addition function May6 comment Generalized addition functionYour solution is correct, I made a mistake in my last edit, there is no problem with the last statement in its current form. May6 revised Generalized addition functiondeleted 393 characters in body May5 revised Generalized addition functionadded 397 characters in body May5 comment Generalized addition function$(-x)\oplus 0= x\oplus 0$ is a desired property May5 comment Generalized addition functionAbout the $0$, $0\oplus 0=(0+0)0\oplus 0=0$. May4 comment Generalized addition functionThanks, I have changed the fourth condition to avoid this inconsistency. May4 revised Generalized addition functionadded 105 characters in body May4 comment Generalized addition functionThanks, I added it in the statement of the problem. May4 revised Generalized addition functionadded 73 characters in body May4 asked Generalized addition function Apr27 comment Deriving the exponential distribution from a shift property of its expectation (equivalent to memorylessness).I don't think $f$ is uniquely determined. It cannot be derived from the differential equation $F(a)+\mu F'(a)=1$ since this equation is not valid on a set of measure zero. At least I don't know how to solve such equations. I agree that the exponential distribution is a solution to the above problem. I have a feeling that it might not be unique and I'd like to have an argument for or against my feeling... Apr26 comment Deriving the exponential distribution from a shift property of its expectation (equivalent to memorylessness).Thank you for replying, I'd love a clarification for my own self. Correct me if I'm wrong, but I thought that the FTOC can only guarantee that the derivative exists almost everywhere. This implies that the first order differential equation is only almost everywhere satisfied by $F(x)$, and therefore is not uniquely given by the exponential solution any more. Apr26 answered A Question on Outer Measure