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16

No. The Euclid-Euler theorem states that any even perfect number $n$ (we don't know whether there are any odd ones) is of the form $$n = 2^{k-1}(2^k - 1)$$ with $2^k - 1$ prime, and furthermore that any $n$ of that form is perfect (this last part is relatively easy to prove, but it is the former part you need). This is clearly not a square, since $2^k - 1$...

11

You likely know that divisibilty by $9$ can be decided via checking if the sum of digits is divisible by $9$. More specifically, it is true that the sum of digits of a number $n$ is congruent to $n$ modulo $9$. So, to show what you want you can also show that that the perfect number itself is congruent to $1$ modulo $9$. To do this write it as $$1 + \... 11 This is Problem 12019 in the American Math Monthly. A solution can be found in the link: AMM 12019 10 As pointed out by E. Schmidt, the sequence A023042 shows that a large percentage of cubes N^3 are a sum of three positive cubes. OEIS lists only N<1770, but we can extend that:$$\begin{array}{|c|c|} \hline N&\text{%}\\ \hline 2000&85.8\text{%}\\ 4000&89.8\text{%}\\ 6000&92.1\text{%}\\ 8000&93.3\text{%}\\ 10000&94.2\text{%}\\ \...

10

A perfect square can be written on the form $n=(p_1^{\alpha_1}\cdots p_k^{\alpha_k})^2$ where $p_i$ are some prime numbers and $\alpha_i> 0$ are integers. The sum of the divisors of $n$ is given by $$\sigma(n) = \prod_{i=1}^k \sigma(p_i^{2\alpha_i}) = \prod_{i=1}^k(1+p_i + \ldots+p_i^{2\alpha_1})$$ The term $1+p_i+\ldots+p_i^{2\alpha_i}$ is an odd ...

10

Any connection with Mersenne numbers is a red herring. We have $\left(\frac {1+\sqrt 5}2\right)^n+ \left(\frac {1-\sqrt 5}2\right)^n$ is exactly an integer for even $n$ because all the terms with odd powers of $\sqrt 5$ cancel. Because $\left|\frac {1-\sqrt 5}2\right|\lt 1$ when we raise it to a large power it becomes very small, so $\left(\frac {1+\sqrt 5}... 8 No it cannot. Let$n=\prod p_i^{\alpha_i}=2^{\alpha_1}3^{\alpha_2}5^{\alpha_3}\cdots$where$p_i$is the$i$th prime number and let$S(n)$be the sum of the divisors of$n$. Suppose$n$is an odd perfect number divisible by$825=3\cdot 5^2\cdot 11$, then$S(n)=2n$and$\alpha_1=0,\alpha_2\ge 1,\alpha_3\ge 2,\alpha_5 \ge 1. Since \begin{align} S(n)&... 8 I did look at the link, but of course the first thing I thought of was Euler's form of an odd perfect number from which this follows immediately. Euler proved that the prime decomposition n=q^\alpha p_1^{2e_1}\cdots p_m^{2e_m} where q\equiv \alpha\equiv 1\operatorname{mod} 4, i.e. a perfect square times one extra prime to a power. Taking this into ... 8 The divisors of p^n are 1,p,p^2,\ldots,p^n. The proper divisors are all but the last one. The sum of those is 1+p+p^2+p^3+\cdots+p^{n-1}. $$This is a geometric series. Apply the standard formula for the sum of a finite geometric series and see if you get p^n. 7 1. By a theorem of Euler, any even perfect number is of the shape 2^{p-1}(2^p-1) where p is prime. We don't need the primality part. Multiply by 8, add 1. We get 2^{2p+2}-2(2^{p+1})+1, which is the square of 2^{p+1}-1. 2. We show that an odd perfect number cannot have only 2 distinct prime factors (we don't deal with only 1 prime factor, ... 7 The theorems are of the form, "If any odd perfect numbers exist, then they must be of the form..." If there are no odd perfect numbers, then these theorems are vacuously true (and indeed, one of the ways of eventually proving there are no odd perfect numbers would be to prove that if any existed, then some contradictory properties would have to be true ... 7 Hint: Find some proper divisors of kn that add up to exactly kn. Are any divisors missing? 7 Being odd or being perfect for some integer n is not at all related to what base we are working over for number notation. (it's the essence of the answer in your linked Q&A as well, so you should have known). So it's still open. n is an odd integer means that n = 2m+1 for some integer m (here with integers I mean \Bbb N=\{0,1,2,3,\ldots\}) (... 6 Suppose m|2n. Then either m|n or \frac{1}{2}m|n. Hence the set of divisors is \{d,2d:d|n\} which has sum at most 3\sum d = 6n. Use the even property to get the strict inequality. Edit: In case this wasn't clear, you know that at least one divisor appears doubly in the list, namely n=2(n/2). Re-edit: To make the argument more succinctly, as ... 6 As Andre mentioned, the even perfect numbers have the form 2^{p-1} (2^{p} -1 ). Claim: The sum of the first 2^{(p-1)/2} odd cubes is equal to this perfect number. Proof: We know that the sum of the first n odd cubes is: \sum_{i=1}^{n} (2i-1)^3 = \sum_{i=1}^{2n} i^3 - \sum_{i=1}^{n} (2i)^3 \\= \left[ \frac {(2n)(2n+1)}{2} \right] ^2 - 8\left[ \frac {... 6 A quick way to prove that no prime power is perfect is to notice that, if p is a prime, then p \geq 2, so that$$\frac{\sigma(p^k)}{p^k} = \frac{1 + p + \ldots + p^k}{p^k} = \frac{p^{k+1} - 1}{p^k(p - 1)} < \frac{p^{k+1}}{p^k(p - 1)} = \frac{p}{p - 1}.$$Now, since p \geq 2, we get that$$\frac{1}{p} \leq \frac{1}{2} \Longrightarrow -\frac{1}{p} \... 6 Let\sigma$denote the sum-of-divisors function. A positive integer$N$is said to be perfect if$\frac{\sigma(N)}{N}=2$. Let$p^\alpha$be an odd prime power and suppose that$N=p^\alpha$is an odd perfect number. Since$\sigma(p^\alpha)=1+p+p^2+\dots+p^\alpha=\frac{p^{\alpha+1}-1}{p-1}, we can observe that \sigma(p^\alpha) = \frac{p^{\alpha+1}-1}{p-1}... 6 Well, let's see. Even perfect numbers are all of the form 2^{n-1}(2^n-1), where 2^n-1 is a Mersenne prime. As such, they are completely determined by the highest power of 2 dividing them. This value in a sum of two such numbers will be the same as in the smallest of them. Really, let N_1=2^{n_1-1}(2^{n_1}-1) < N_2=2^{n_2-1}(2^{n_2}-1),\;n_1<n_2. ... 6 See : Thomas Heath, A History of Greek Mathematics : Vol.I (1921), page 69 : Aristotle observes that the One is reasonably regarded as not being itself a number, because a measure is not the thing measured, but the measure or the One is the beginning (or principle) of number. See also Euclid's Elements, Book VII, Defs.1&2 : A unit is that by ... 6 In fact, to generalize on @winther's result, note that \sigma (n) is odd iff n is of the form k^2 or 2k^2. The proof of this can be found in this post. Thus, since for a perfect number \sigma(n)=2n \equiv 0 \pmod 2, a perfect number can neither be of the form k^2 or 2k^2. Thus, any square-or any number that is twice a square-is not a ... 6 As tomi pointed out in the comments, there's nothing about the sequence that you need to keep in order, so it's more basic to just talk about sets. Any Egyptian fraction sequence can be indefinitely prolonged, so we can write, for instance, \begin{align} 2 & = 1+\frac{1}{2}+\frac{1}{2} \\ & = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{6} \\ & = 1+\... 6 You're wrong. All that Euclid proved was that when 2^p-1 is prime, then 2^{p-1}(2^p-1) is perfect. He never said that all perfect numbers can be obtained by this process. In the XVIIIth century, Euler proved that every even perfect number can be obtained by that process. He never stated that this holds for odd perfect numbers too. 6 It's "the sphere", or "any sphere", not "a sphere". Most likely you observed that 7 is a Mersenne prime and 28 is the associated perfect number. There's indeed a connection, though maybe not exactly what you expect: the number of smooth structures on the (4k-1)-sphere is divisible by 2^{2k-2}(2^{2k-1}-1); see wikipedia. When k=2 you happen to have 4k-... 6 Maybe it is easy to see in base10:999990000 = 99999\times 10^4=(10^5-1)10^4$$In base2:$$(111110000)_2 = (11111)_2\times 2^4=(2^5-1)2^4$$If you're not comfortable with base2, change it to base10:$$\begin{align}(111110000)_2 &= 0+0+0+0+2^4+2^5+2^6+2^7+2^8 \\ &= 2^4(1+2+2^2+2^3+2^4) \\ &= 2^4(2^5-1) \end{align}$$5 In mathematics, to claim (A) All unicorns are mammals. is exactly the same as claiming (B) There is nothing that is a unicorn but not a mammal. One way this can be true is that if there are no unicorns at all. A universal statement such as (A) is "true by default": the only way it can be false is if there is a counterexample to it. A counterexample ... 5 The Mersenne Twister  is a good pseudorandom number generator based on a Mersenne prime. Apparently other systems exist using Mersenne primes for pseudorandom number generation as well.  In communication complexity, Mersenne primes enabled a major advance in private information retrieval schemes , with an asymptotic result dependent on their ... 5 Let N = {q^k}{n^2} be an odd perfect number given in Eulerian form. Let$$I(x) = \dfrac{\sigma(x)}{x}$be the abundancy index of$x$. This is a partial answer to the original question, and proves the claim in the affirmative, subject to the validity of a recent proof claim by Patrick A. Brown that$q^k < n$holds (in many cases). First, we show the ... 5 Since$\sigma(n)>0$we have$d<2N$, together with$(2/3)N^2<d$we get$(2/3)N^2<2N$which is verified only for$0< N< 3$, which leaves only 2 cases to be checked, namely$1$and$2$. edit:After the question being asked has been changed the first part of my answer is irrelevant. I can't completely answer the new question, but I can show ... 5 For a counterexample, look at$n=945=3^3\cdot 5\cdot 7$. It turns out that$\sigma(n)=1920\gt 1890$. There are infinitely many other counterexamples. One family is obtained by multiplying$945$by any odd number. There are other families. 5 No, that's false. The numbers for which$\sigma(n) = 2n$are called perfect, while$\sigma(n) > 2n$are called abundant. It is currently unknown whether there are odd perfect numbers or not, while the smallest odd abundant number is$945\$. Even more: there are infinitely many odd abundant numbers (cfr. OEIS sequence A005231). This is because every ...

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