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Denote by $B_r$ the $r$-th Bernoulli polynomial. Are there any positive integers $r, x$ such that. $B_r(x)$ divides $B_r(x+1)$ or vice versa ?

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We know that $B_n(x)- B_n\in \mathbb Z$, so for even $n$ and integer $x$ number $B_n(x)$ is not an integer.

Let $n$ be odd. In this case $B_n=0$ and $$B_n(x)=n(1^{n-1}+\cdots+ (x-1)^{n-1}).$$ If $B_n(x)\mid B_{n+1}(x)$ then from formula $B_n(x+1)-B_n(x)=nx^{n-1}$ follows that $B_n(x)\mid nx^{n-1}$ i.e. $$1^{n-1}+\cdots+ (x-1)^{n-1}\mid x^{n-1}.$$ After this step you'll come to the question similar to Erdős–Moser equation.

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