# André Nicolas

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 15h comment For any Fermat number $F_n=2^{2^n}+1$ with $n>0$, establish that $F_n\equiv5\,\text{or}\,8\pmod9$ Yes, thank you, fixed. 15h revised For any Fermat number $F_n=2^{2^n}+1$ with $n>0$, establish that $F_n\equiv5\,\text{or}\,8\pmod9$ edited body 17h comment Calculate variance If one makes the assumption of independence, the problem is fairly well specified. One can poke fun at the independence assumption. We then have a binomial $n$ large $p$ small, $np=5$ medium. Standard setup for using the Poisson approximation to the binomial. 17h comment Calculate variance Call a computer crashing a success. Probability $p$ of success is $\frac{1}{800}$, number $n$ of independent (?) trials (computers) is $4000$. 17h comment Calculate variance You may be expected to use the Poisson approximation, variance is $\lambda$. But binomial is better. We are assuming independence, which seems physically unreasonable. 17h revised Find the maximum angle possible added 24 characters in body 17h comment Looking for Proofs on Finding the Roots of an Algebraic Congruence with a Composite Modulus The fact we get all solutions follows from the fact that if $a\equiv a'\pmod{m}$ and $b\equiv b'\pmod{p}$, then $a+b\equiv a'+b'\pmod{m}$ and $ab\equiv a'b'\pmod{m}$. Polynomials are built up using addition and multiplication. As to positive residues or least absolute residues, the results will be the same (modulo $m$). Least absolute residues can be useful because the arithmetic may be easier, and one can take advantage of $\pm$ symmetries. 17h revised For any Fermat number $F_n=2^{2^n}+1$ with $n>0$, establish that $F_n\equiv5\,\text{or}\,8\pmod9$ added 154 characters in body 17h answered For any Fermat number $F_n=2^{2^n}+1$ with $n>0$, establish that $F_n\equiv5\,\text{or}\,8\pmod9$ 18h answered Find the maximum angle possible 18h answered Convergence/ divergence of a series 18h comment cardinality of the set of $\varphi: \mathbb N \to \mathbb N$ such that $\varphi$ is an increasing sequence (Cont) So the set of increasing sequences has cardinality $\ge c$. But the set of increasing sequences also has cardinality $\le c$. Therefore the vardinality is exactly $c$. 18h comment cardinality of the set of $\varphi: \mathbb N \to \mathbb N$ such that $\varphi$ is an increasing sequence @GinKin: From any binary sequence, we produce an increasing sequence of natural numbers. It is clear that two distinct binary sequences produce distinct increasing sequences. So we have produced an injective mapping from binary sequences to increasing sequences. To put it another way, the set of binary sequences has the same cardinality as a subset of the increasing sequences. Thus the cardinality of the set of increasing sequences is at least as big as the cardinality of the set of binary sequences. The set of binary sequences has cardinality $c$. (Cont) 1d revised Tossing Dice repeatedly, probability that 2nd trial had more tosses than 1st one added 285 characters in body 1d comment Tossing Dice repeatedly, probability that 2nd trial had more tosses than 1st one The infinite geometric sequence I referred to has sum $\frac{1}{36}\frac{1}{1-\frac{25}{36}}=\frac{1}{11}$. Thus $a=\frac{5}{11}$. 1d comment Finding the remainder of a linear congruence It has been done on MSE too many times. 1d revised Tossing Dice repeatedly, probability that 2nd trial had more tosses than 1st one added 75 characters in body 1d comment Finding the remainder of a linear congruence Too lengthy for a comment. Editing comments is very unpleasant, there is a $5$-minute clock. 1d revised Tossing Dice repeatedly, probability that 2nd trial had more tosses than 1st one added 57 characters in body 1d answered Tossing Dice repeatedly, probability that 2nd trial had more tosses than 1st one