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I will show that even the product of the first $n-1$ terms of the set is not divisible by the $n$'th term; this is clearly enough since it has to go to one of the subsets. If $n = 1, 2, \ldots , 6$, t …

I apologize for posting another answer, but apparently everyone, including me, missed a very simple proof why the product is never a square, which is indeed suitable for children: just consider everyt …

We will ignore most of your assumptions, let $d = d_1$ be a number and assume that for $1 \le k \le x$ we have that $k$ is a square modulo $d$ if and only if $k$ is a square in $\mathbb{N}$. We will s …

Even one such factor gives you zero density. Indeed, if $d \mid n$, $\alpha \sqrt{n} \le d \le \beta \sqrt{n}$ then $\frac{1}{\beta}\sqrt{n} \le \frac{n}{d} \le \frac{1}{\alpha}\sqrt{n}$ therefore $n$ …

I will hopefully prove this statement below modulo a finite check (I believe this will also work to show that $\gcd(\binom{b}{3}, \binom{a}{3}) \le \varepsilon b(a-b)^3$ for any $\varepsilon > 0$ exce …

I have no idea if it is known, but here is the proof: First of all, we can remove the coprimality condition by factoring out the gcd and just prove that $S=\sum_{a, b = 1}^\infty \frac{1}{ab(a+b)} = 2 …

I will prove it modulo a fact written in Wikipedia which I don't know how to prove (but since, as we all know, Wikipedia is the universal arbiter of truth and is never wrong, I will equate it with com …

Let $a\in S$ be some fixed element. Note that $a\oplus b \le a + b$. Let $N$ be some big number. Put $M = [1, \ldots , N]\cap S$. We have $a\oplus M \cap M = \varnothing$. We also have $a\oplus M \sub …

I will prove below that your bound $\frac{\exp\left(C\frac{\log N}{\log \log N}\right)}{A^3}$ (which follows from $\sum_{d\mid N, d > A} \frac{1}{d^3} \le \frac{d(N)}{A^3}$) is optimal at least in the …

UPDATE Since I thought $\sigma_n$ was just the sum of the divisors of $n$ and as @Ella pointed out I was off by a factor of exactly $24$ what is written below is only the (sketch) of the proof of the …

Sure, for each fixed $k$ both of them are (up to a constant) eventually dominated by $c(n)=c(n-1)+c(n-k)$ and this sequence is $O(t^n)$ for $t=t(k)\to 1$ as $k\to \infty$ (one has to prove a simple es …

If I'm not mistaken, it is true that $S$ must contain all primes. First of all it is obvious that $S$ is infinite -- indeed, as Euclid teach us, if $S$ is finite then $\left(\prod\limits_{p\in S} p\r …

To get a bound which is worse than the one of GH from MO asymptotically, but which doesn't require any case checking, we can do the following: if $\gcd(t, N) = k$, then $\frac{N}{k} = d$ which is an i …

Although Iosif Pinelis already gave a slick answer, let me indicate an approach based on a more general theory of entire functions, though I will omit some details to not turn this answer into a liter …

Assume for the sake of contradiction that $a_i = 6, i \in [n, 2n]$. Consider the rational number $$r = \sum_{i = 0}^{n-1} a_i7^{-i} + \frac{1}{7^{n-1}}.$$ Clearly, $r > \sqrt{7}$. On the other hand, …