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1d
comment What's wrong in this proof of $10$ is a solitary number?
@the4seasons Interesting! So if we call the friendly number candidate $5x$, and if we assume $\gcd(5,x)=1$, then the we are led to and impossible equation. Therefore the asker actually proves that no friend of $10$ can be of the form $5x$ with $\gcd(5,x)=1$. Because if that $x$ existed, it would be a friend of $2$, and $2$ is known to be a solitary number. Agree?
1d
comment What's wrong in this proof of $10$ is a solitary number?
What middle part of the proof are you referring to? The asker seems to agree that $\frac{\sigma(x)}{x}>1$ for all $x>1$. However, the first error is that the proof starts by considering a wrong equation at the very beginning.
1d
comment What's wrong in this proof of $10$ is a solitary number?
@the4seasons Yes, I can see how he comes from $\sigma(5x)=9x$ to the next line $6\sigma(x)=9x$ under his assumption. However, the error is before that. There is no reason to regard $\sigma(5x)=9x$ in the first place. Where does that come from?! Bhaskar Vashishth asked the same in an immediate comment to the question. From my post above it should be clear that we should solve $5\sigma(x)=9x$. So his "5" starts under the parenthesis (in the argument to $\sigma$) for no good reason! After that, it becomes a "6", but that is not the error.
May
8
comment How do I square a logarithm?
You: Actually, the only way that $(\log_2(3))^2 = 2 \log_2(3)$ could hold is if $\log_2(3)$ were equal to 2. Maybe that should be "2 or 0" (the solutions to $x^2=2x$). For example, $(\log_2(1))^2 = 2 \log_2(1)$.
May
8
comment Where does the constant increase by 2 of differences between integer square values come from?
@jeremyradcliff There are many related illustrations on the web, see e.g. http://www.math.upenn.edu/~deturck/probsolv/LP1ans.html.
May
6
comment If three complex numbers $z_k$ have modulus $1$, then $|z_1+z_2+z_3| = |\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}|$
The geometric interpretation of your first equation is that as long as $z_i$ is a unit vector, taking the reciprocal of $z_i$ is exactly the same as reflecting $z_i$ in the real line (horizontal axis). So if you imagine three unit vectors given in arbitrary position, with their sum vector constructed, just take the mirror image with respect to the real axis. That will not change the length of the sum vector. I guess this is what user238209 said concisely in his answer, the map $z\mapsto\frac{1}{z}$ is an isometry on the unit circle.
May
5
comment Example of uncomputable but definable number
@MattSamuel Even more generally, every algebraic number (which includes every rational number) is computable.
May
4
comment Result of solving an unsolved problem?
If someone proved $P \lneqq NP$, while it would be of great theoretical and technical importance, would the society in general change much?
Apr
25
comment How Deficient a Number is? (Finding numbers having a certain deficiency)
@JrAntalan Arter you made the edit, it is a new problem. This time it is not hard to find an N that works, like $N=11\cdot 13$.
Apr
25
comment How Deficient a Number is? (Finding numbers having a certain deficiency)
One can use the inequality $\frac23 N^2 < N^2-N$ from the question, directly to eliminate those two cases, so again we do not need to actually calculate any $\sigma$ value.
Apr
25
comment How Deficient a Number is? (Finding numbers having a certain deficiency)
@JrAntalan You need to verify for yourself that neither $1$ nor $2$ satisfy this.
Apr
25
comment Is it possible for integer square roots to add up to another?
In the example $x=3,y=12$ one of the radicands, namely $y=12$, is not square-free. For that reason many people prefer to not write $\sqrt{12}$ and use $2\sqrt{3}$ instead. The relation becomes $\sqrt{3}+2\sqrt{3}=3\sqrt{3}$ which looks less fascinating. If we require that both radicands $x$ and $y$ are square-free, the only way the criterion of this answer can be satisfied is indeed $x=y$, so @Tyg13 becomes right in that sense. And $\sqrt{x}+\sqrt{x}=\sqrt{4x}=2\sqrt{x}$ is again dull. So: If you pull out all square factors so what remains under the radical signs is square-free, no surprises.
Apr
22
comment Does A5 have a subgroup of order 6?
If you want to cheat, you can look at Twisted S3 in A5 and the pages it links at the top of the page.
Apr
17
revised Is $2^{\aleph_0} = \aleph_1$?
added 384 characters in body
Apr
17
answered Is $2^{\aleph_0} = \aleph_1$?
Apr
17
comment A number $n$ which is the sum of all numbers $k$ with $\sigma(k)=n$?
Not sure I understand all you say. But if you want to prove a total amicable tuple cannot exist for $m>1$, be sure to use the "totality" in a crucial way. Because plenty of amicable tuples are known. If for example you could prove that for every non-trivial amicable tuple there exist a "new" integer which is not in the tuple, and whose $\sigma$ value is also $n$, it would suffice. Is there a way to construct such a witness for non-totality?
Apr
17
comment A number $n$ which is the sum of all numbers $k$ with $\sigma(k)=n$?
Since neither of the two late "answers" really gives the answer, I shall award the bounty to this one.
Apr
16
comment A number $n$ which is the sum of all numbers $k$ with $\sigma(k)=n$?
Similarly, for an amicable triple like $(k_1,k_2,k_3)=(1980, 2016, 2556)$ with $k_1<k_2<k_3$ we have $I(k_1)>I(k_2)>I(k_3)$. So here on (harmonic) average the abundancy of the three $k_i$ is $3$. So we could have that all three are abundant. (Or maybe the first two, $k_1$ and $k_2$, are very abundant, and the last one, $k_3$, is perfect/deficient, in a way so that the average abundancy I describe is still $3$.) Again this example triple is not "total". For an example of an amicable (not total) quadruple ("average" abundancy is $4$), check $(k_1,k_2,k_3,k_4)=(3270960,3361680,3461040,3834000)$.
Apr
16
comment A number $n$ which is the sum of all numbers $k$ with $\sigma(k)=n$?
What you denote by $I(k)$ I call the abundancy (index) of $k$. If $(k_1,k_2)$ is an amicable pair, and we agree that it is sorted ascendingly by $k$, so $k_1<k_2$, then clearly the abundancy is descending, $I(k_1)>I(k_2)$. Since $k_1+k_2=n$, we have $\frac{k_1}{n}+\frac{k_2}{n}=1$ or $\frac12 ( \frac{k_1}{n}+\frac{k_2}{n} ) =\frac12$, so on (harmonic) average the abundancy $\frac{n}{k}$ is $2$. So $k_1$ is abundant and $k_2$ is deficient. Example: $(k_1,k_2)=(220,284)$. However this pair is not "total" by my definition from the question.
Apr
16
comment Caden has 4/3 kg of sand which fills 2/3 ​​ of his bucket. How many buckets will 1kg sand fill?
This is a division question, specifically $\frac{\frac23}{\frac43} = \frac12$.