This is a problem about Mathematical Analysis. If there exist infinite sets, select a set $T=\{p_1,p_2, \cdots, p_t\}$, $t$ is bounded, and $p_t$ is lagre enough, $p_1<p_2<\cdots<p_t$. there exist $1 \leq i \leq t$, $p_t \mid p_i+k $, $i \neq t$ because $p_t>k$, then $p_t \leq p_i+k \leq p_{t-1}+k$, $p_{t-1} \geq p_t-k$ is also lagre enough. Now, considering $p_{t-1} \mid p_i+k$, $i \neq t-1$ because $p_{t-1}$ is lagre enough. If $i=t$, $p_{t-1} \leq \frac{p_t+k}{2}<p_t-k \leq p_{t-1}$, a contradiction. So we get $p_{t-1} \leq p_i+k \leq p_{t-2}+k$, $p_{t-2} \geq p_t-2k$. We can get $p_i \geq p_t-k(t-i)$ are all lagre enough. $p_1 \cdots p_t \mid (p_1+k)\cdots (p_t+k)-p_1 \cdots p_t=O(p_t^{t-1})$, but $p_1 \cdots p_t>(p_t-kt)^t>O(p_t^{t-1})$ when $p_t$ is lagre enough, a contradiction.

WypHxr wrote: This is a problem about Mathematical Analysis. If there exist infinite sets, select a set $T=\{p_1,p_2, \cdots, p_t\}$, $t$ is bounded, and $p_t$ is lagre enough, $p_1<p_2<\cdots<p_t$. there exist $1 \leq i \leq t$, $p_t \mid p_i+k $, $i \neq t$ because $p_t>k$, then $p_t \leq p_i+k \leq p_{t-1}+k$, $p_{t-1} \geq p_t-k$ is also lagre enough. Now, considering $p_{t-1} \mid p_i+k$, $i \neq t-1$ because $p_{t-1}$ is lagre enough. If $i=t$, $p_{t-1} \leq \frac{p_t+k}{2}<p_t-k \leq p_{t-1}$, a contradiction. So we get $p_{t-1} \leq p_i+k \leq p_{t-2}+k$, $p_{t-2} \geq p_t-2k$. We can get $p_i \geq p_t-k(t-i)$ are all lagre enough. $p_1 \cdots p_t \mid (p_1+k)\cdots (p_t+k)-p_1 \cdots p_t=O(p_t^{t-1})$, but $p_1 \cdots p_t>(p_t-kt)^t>O(p_t^{t-1})$ when $p_t$ is lagre enough, a contradiction. I don't think your solution makes sense, as you claimed "large enough" ambiguously. To be precise, to get $p_2 | p_1+k$, as you said, $p_3 >3k$ is required. Succesively, you need $p_t >tk$, but the restriction is related to $t$ which varies.

liekkas wrote: WypHxr wrote: This is a problem about Mathematical Analysis. If there exist infinite sets, select a set $T=\{p_1,p_2, \cdots, p_t\}$, $t$ is bounded, and $p_t$ is lagre enough, $p_1<p_2<\cdots<p_t$. there exist $1 \leq i \leq t$, $p_t \mid p_i+k $, $i \neq t$ because $p_t>k$, then $p_t \leq p_i+k \leq p_{t-1}+k$, $p_{t-1} \geq p_t-k$ is also lagre enough. Now, considering $p_{t-1} \mid p_i+k$, $i \neq t-1$ because $p_{t-1}$ is lagre enough. If $i=t$, $p_{t-1} \leq \frac{p_t+k}{2}<p_t-k \leq p_{t-1}$, a contradiction. So we get $p_{t-1} \leq p_i+k \leq p_{t-2}+k$, $p_{t-2} \geq p_t-2k$. We can get $p_i \geq p_t-k(t-i)$ are all lagre enough. $p_1 \cdots p_t \mid (p_1+k)\cdots (p_t+k)-p_1 \cdots p_t=O(p_t^{t-1})$, but $p_1 \cdots p_t>(p_t-kt)^t>O(p_t^{t-1})$ when $p_t$ is lagre enough, a contradiction. I don't think your solution makes sense, as you claimed "large enough" ambiguously. To be precise, to get $p_2 | p_1+k$, as you said, $p_3 >3k$ is required. Succesively, you need $p_t >tk$, but the restriction is related to $t$ which varies. my proof is wrong, because $t$ can also lagre enough, we can use $p_t \geq t \ln t$ to do it

My proof is wrong, because $t$ can be also lagre enough, we can use $p_t \geq t \ln t$ to do it

Let $S$ be a set satisfying the properties and $x$ be the largest element in $S$. It suffices to show $x$ is bounded (then there are finitely many sets containing only elements less than our bound). Indeed, let $q$ be the smallest prime not dividing $k$ and suppose $x > 3qk$. Clearly $p+k < 2x \forall p \in S$, so $x-k \in S$. Similarly, we obtain $x-2k, x-3k, \cdots x-qk \in S$ (since $p+k \leq x+k < 2(x-qk)$). But by Pigeonhole Principle, one of those elements is a multiple of $q$ (and not $q$), a contradiction. The conclusion follows.

Henry_2001 wrote: Prove that for any positive integer $k,$ there exist finite sets $T$ satisfying the following two properties: $(1)T$ consists of finite prime numbers; $(2)\textup{ }\prod_{p\in T} (p+k)$ is divisible by $ \prod_{p\in T} p.$ For $k=1$, choose $T=\{2,3\}$. Then $(2+1)(3+1)=12$ is divisible by $2\cdot3=6$. For $k\ge2$, let $p$ be a prime divisor of $k$ so that $k=pm$. Choose $T=\{p\}$. Then $p+k=p(1+m)$ is divisible by $p$.

Henry_2001 wrote: Prove that for any positive integer $k,$ there exist finite sets $T$ satisfying the following two properties: $(1)T$ consists of finite prime numbers; $(2)\textup{ }\prod_{p\in T} (p+k)$ is divisible by $ \prod_{p\in T} p.$ The translation should be: ... there exist finitely many sets ... and ... consists of finitely many prime numbers.

sccdgsy wrote: Henry_2001 wrote: Prove that for any positive integer $k,$ there exist finite sets $T$ satisfying the following two properties: $(1)T$ consists of finite prime numbers; $(2)\textup{ }\prod_{p\in T} (p+k)$ is divisible by $ \prod_{p\in T} p.$ The translation should be: ... there exist finitely many sets ... and ... consists of finitely many prime numbers. Thank you.

Let's see if someone can spot the mistake(if there is one) )) FTSOC suppose that there are infinitely such sets $T$. Since there are infinitely many $T$, we can choose $T_0$ such that $p_l = max(T_0) > t \cdot k$ for some $t \in \mathbb{N}$ (we will choose it later) $p_l | \prod_{p \in T_0} (p + k) \implies$ there is index $i$, such that $p_l | p_i + k$ Note that $p_i + k < p_l + k < p_l + t \cdot k < p_l + p_l = 2 \cdot p_l \implies p_i + k = p_l$ Note that $p_i = p_k - k > (t - 1) \cdot k$ Since $p_i | \prod (p + k) \implies \exists j: p_i | p_j + k$ If $p_j > p_i \implies p_i < p_j < p_l = p_i + k \implies p_j + k < p_i + k + k < p_i + 2k > p_i + (t - 1) \cdot k < p_i + p_i = 2 \cdot p_i \implies p_j + k = p_i$, a contradiction Then $p_j < p_i \implies p_j + k < p_i + k < p_i + (t - 1) \cdot k < 2 \cdot p_i \implies p_j + k = p_i$ Observe that such manipulations could be repeated for $p_j$ Then, we can choose a sequence $p_1, p_2, \ldots p_{t - 1}$ , from $T_0$, for which the following condition holds: $p_{i - 1} + k = p_i \forall i = 2, 3, \ldots t - 1$ So the sequence looks like $x, x + k, x + 2k, \ldots x + (t - 2)k$ Lets take $t = 2k + 3$, so in the sequence there is at least two zeroes modulo $k$, a contradiction because all terms of sequence must be primes, so there cannot be infinitely many such sets