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Can one partition the set of positive integers into finitely many Pythagorean-triple-free subsets? If so, what is the smallest number of such subsets? Taking a wild guess, I would be least surprised …

If the Collatz conjecture is strongly false, in the sense that there is an infinite orbit, let $S_n$ be the set of natural numbers $\le n$ whose orbit goes to infinity. If $c=\liminf _{n\rightarrow\in …

Can one color the positive integers with finitely many colors, so that no two different numbers of the same color add to a square? Some easy to prove remarks: at least 4 colors are needed, since the …

Define the Pythagorean Graph as having nodes $a,b\in \mathbb{N}_{\ge 3}$ and an edge $a\rightarrow b$ if and only if $a^2+b^2$ is a square. After much searching I found the example in the picture, pro …

The answer is no. Start with $\alpha$ between 3.25 and 3.75; take squares of those 2 to see that one can restrict $\alpha^2$ to being between 11.25 and 11.75; take 3/2 powers of those to see that one …

Denoting by $p_i$ the $i$-th prime, is it known that $\displaystyle \sum_{i=1}^\infty \frac{1}{p_{i+1}^2-p_i^2}$ converges? Can one compute a few digits based on euristic considerations or plausible c …

Define $\text{rad}_{23}(2^m3^nr)=2^{\text{sign}(m)}3^{\text{sign}(n)}r$, where $m,n\ge0$ and $2,3\nmid r\in\mathbb{N}$. For a triple $a+b=c$ define the quality $q_{23}(a,b,c)=\frac{\log(c)}{\log(\tex …

Define a set of numbers with small radicals (A341645 in OEIS) by $$A_2=\{n\in\mathbb{N} \;|\; \text{rad}(n)^2\le n\}$$ The asymptotic density of $A_2\cap \{1,\dots N\}$ is $\sqrt{N}\times e^{2(1+o(1)) …

If $\{p_i\}$ is the sequence of all primes, is it possible that there exist a non constant $P\in \mathbb{Z}[x_1,\dots x_n]$ such that $P(p_i,p_{i+1},\dots p_{i+n-1})$ is bounded in $i$? More precisely …

Modulo 6 the squares are 0,1,4,3,4,1 and the squares+7 (or -5) can only be 1,2,5,4,5,2, of which 3/6 can at all be prime. The squares-7 (or +5) are 5,0,3,2,3,0 of which only 1/6 can be prime. Obviousl …

Not an answer - but I decided to delete a prior comment and repost as an answer, because I think it puts the 2-4-6-8 conjecture in a different light than considered so far, hopefully leading to some o …

The first two equalities imply $x>m$ and $y>n$ so one can substitute $x=m+X$, $y=n+Y$ and $k=X+Y$, with still $X,Y \in \mathbb N$: ${X \choose 2}=nX+nY-mX\tag{1}$ ${Y \choose 2}=mX+mY-nY\tag{2}$ Fr …

Here is a completely elementary proof, inspired by Pasten's comments. Let $P(n)=an^2+bn+c$. Take $n=a^5x^4+2a^3(ab+2a+1)x^3+a(2a^3c+a^2b^2+6a^2b+3ab+6a^2+5a+1)x^2+(ab+2a+1)(2a^2c+2ab+b+2a+1)x+a^3c^2+2 …

I suspect the answer is no. First rewrite $x_n=y_n+1$, then the recursion becomes $(n+1)y_{n+1}=ny_n(y_n+2)=(y_n+2)(y_{n-1}+2)\cdots (y_2+2) (y_1+2)y_1$ and for the integrality of $y_{n+1}$ it is su …

Define the size, possibly $\infty$, of a set $S\subseteq \mathbb{N}$ as $|S|=\sum\limits_{n\in S} \frac{1}{n}$. Then the Erdős-Turan conjecture states that if $|S|=\infty$, S must contain arbitrarily …