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I would like to ask the following question: Is it possible to find two non-constant polynomials $p(x), q(x)$ with integer coefficients, such that $\gcd(p(n), q(m))=1$ for every $(n, m)\in \mathbb{N} …

Fermat proved that $x^3-y^2=2$ has only one solution $(x,y)=(3,5)$. After some search, I only found proofs using factorization over the ring $Z[\sqrt{-2}]$. My question is: Is this Fermat's original p …

Let $p$ be a prime number and $P=\{1,2,...,p-1\}$ In how many ways we can sum all the elements of $P$ in such a way that we will reach a multiple of $p$ only when we sum the last summand? For …

Let $P$ be a polynomial having integer coefficients (and degree $\geq 3$), and let $\mathscr P_P$ be the set of prime numbers dividing some value $P(n)$ with $n \in \mathbb Z$. Is it true that $\ …

Sierpsinski proved that whenever this diophantine equation does not have a solution for $x$ then $x^2+(x+1)^2$ is prime. It is conjectured also by Sierpinski that there are infinitely many primes of t …

Suppose we start with $k$ primes $p_1,p_2,\ldots ,p_k$ (not necessarily consecutive) and a residue class for each prime $r_1,r_2,\ldots ,r_k$. We denote the least integer not covered by the arithmetic …

Erdos mentions in his book "Topics in the Theory of Numbers" the following: "It is stil unkown if the equation $\omega(n+1)=\omega(n)$ has infinitely many solutions...It is known that $|\omega …

It is known that the only numbers not "consecutive summable" are the powers of $2$. This is easy to prove: If $m+m+1+\ldots m+k=2^a$ then $2^a=\frac{(k+1)(2m+k)}{2}$. This means that $2^{a+1}=(k+1)(2m …

Let $N=\{1,2,3,\ldots, n\}$. We sum all the elements of every nonempty subset of $N$. Which sum(s) appears most often? (Let's call this sum a champion). Using a simple pigeonhole argument a champion m …

It is well known that $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$ and this is the only solution to $\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}=1$ with $2\leq x_1<x_2<x_3$. My question is: Let $n\in …

I think the question in the title is clear. Let $n\in \mathbb{N}$. It is a nice exercise to show that every prime number divides infinitely many terms of the sequence $2^n-n$. (For example take $n=(p- …

Some days ago, I noticed that $3\cdot 5\cdot 7\cdot 11 +1=34^2$. I am almost sure that if we denote four consecutive primes by $p, q, r, s$ then the equation $$p\cdot q\cdot r\cdot s+1=x^2 \qua …

From Fermat's little theorem we know that every odd prime $p$ divides $2^a-1$ with $a=p-1$. Is it possible to prove that there are infinitely many primes not dividing $2^a+2^b-1$? (With $2^a …

I know that there are no solutions to $2^n\equiv 1\pmod{n}$ for $n>1$ and I can prove that there are infinitely many $n$ such that $2^{n+1}\equiv1\pmod{n}$. My question is: Do we know other fixe …

Let $m$ be an odd integer greater than $1$. Is it true that there are positive $a, b$ such that $m=a+b$ and $a^2+b^2$ is a prime number? It seems that for every odd $m$ there are many $(a,b)\in \mathb …