Either I (*coughs* Google Translate) misunderstood this problem, or it is too easy. Note that $a_n-p_n=a_1-p_1=c \ge 0$. If $c=0$ we're done. Else $p_n \mid (a_n-p_n)$ hence $p_n \le c$ so $a_n \le 2c$. This fails for $n$ too big.

It’s indeed easy, but I’m not surprised since it is Grade 9 P1.

Define $b_n=\frac{a_n}{p_n}$. So, $a_n-a_k=p_n.b_n-p_k.b_k=p_n-p_k\Rightarrow p_n(b_n-1)=p_k(b_k-1)\Rightarrow p_k|b_n-1\forall k>0$& fixed $n$. So,$b_n-1$ has infinitely many prime divisors$\Rightarrow b_n-1=0\Rightarrow b_n=1.$ As the $n$ was chosen arbitrarily, hence this is true for all $n>0$and hence $a_n=p_n\forall n$.

dgmath97 wrote: Define $b_n=\frac{a_n}{p_n}$. So, $a_n-a_k=p_n.b_n-p_k.b_k=p_n-p_k\Rightarrow p_n(b_n-1)=p_k(b_k-1)\Rightarrow p_k|b_n-1\forall k>0$& fixed $n$. So,$b_n-1$ has infinitely many prime divisors$\Rightarrow b_n-1=0\Rightarrow b_n=1.$ As the $n$ was chosen arbitrarily, hence this is true for all $n>0$and hence $a_n=p_n\forall n$. But why $b_n-1$ has infinitely many prime divisors?

Another simple proof: $a_n-a_k=p_n-p_k \to a_n-p_n=a_k-p_k=c$ $\forall p_n, \frac{a_n}{p_n}-1=\frac{c}{p_n} \in \mathbb{Z}$ and because it's either infinity or zero for c to exist, it must be 0, so $a_n=p_n$

Joy_Gomuj wrote: dgmath97 wrote: Define $b_n=\frac{a_n}{p_n}$. So, $a_n-a_k=p_n.b_n-p_k.b_k=p_n-p_k\Rightarrow p_n(b_n-1)=p_k(b_k-1)\Rightarrow p_k|b_n-1\forall k>0$& fixed $n$. So,$b_n-1$ has infinitely many prime divisors$\Rightarrow b_n-1=0\Rightarrow b_n=1.$ As the $n$ was chosen arbitrarily, hence this is true for all $n>0$and hence $a_n=p_n\forall n$. But why $b_n-1$ has infinitely many prime divisors? because for every arbitrary k, the fact that $p_k|b_n-1$ holds

chiyukiRIP wrote: Joy_Gomuj wrote: dgmath97 wrote: Define $b_n=\frac{a_n}{p_n}$. So, $a_n-a_k=p_n.b_n-p_k.b_k=p_n-p_k\Rightarrow p_n(b_n-1)=p_k(b_k-1)\Rightarrow p_k|b_n-1\forall k>0$& fixed $n$. So,$b_n-1$ has infinitely many prime divisors$\Rightarrow b_n-1=0\Rightarrow b_n=1.$ As the $n$ was chosen arbitrarily, hence this is true for all $n>0$and hence $a_n=p_n\forall n$. But why $b_n-1$ has infinitely many prime divisors? because for every arbitrary k, the fact that $p_k|b_n-1$ holds ok I assumed that $k$ must be less than $n$

Another way of seeing this is through letting $a_n=p_n b_n$; $a_n-p_n = p_n(b_n-1)$ is fixed (and is equal to $p_1(b_1-1)$). Pass to a subsequence $p_{n_k}$ such that $p_{n_k}\to \infty$; we get a contradiction unless $b_{n_k}=1$ for sufficiently large $k$, and hence $b_{n}=1$ for every $n$, meaning $a_n=p_n$.

Let $a_{n} = p_{n}b_{n}$. The equation $a_{n}-a_{k} = p_{n}-p_{k}$ is equivalent to $p_{n}(b_{n}-1) = p_{k}(b_{k}-1)$. If $b_{k}=1$ for some $k$ then it is obvious that $b_{n} = 1$ for all $n$ and then $a_{n}=p_{n}$ so the claim holds. Assume now that there does not exist $k$ such that $b_{k} = 1$. Then $b_{i} > 1$ for all $i \in \mathbb{N}$. Since $p_{i+1} > p_{i}$ (this follows from $p_{i+1}-p_{i} = a_{i+1}-a_{i}$ and the assumption that $a_{1}, a_{2}, \dots$ is strictly increasing) the equation $p_{i+1}(b_{i+1}-1) = p_{i}(b_{i}-1)$ implies $b_{i+1}-1 < b_{i}-1$. Hence $b_{1}, b_{2}, \dots $ is strictly decreasing. As this sequence consists only of positive integers then eventually $b_{m}=1$ for some $m$, a contradiction.

What is meant by $(a_n) _n$

hydrohelium wrote: What is meant by $(a_n) _n$ It denotes the sequence for $n$ terms and the $a_n$ denotes the elements of the sequence.

Trivial Though almost same as above but still posting for sake of doing Define a function $s_n$ such that $a_n = p_n \cdot s_n $ for all $n \in \mathbb{N}$.Since $p_n | a_n$ so this function takes positive integral values. Now using the second condition we get $a_n - a_k = s_n p_n - s_k p_k = p_n - p_k \implies p_n (s_n - 1) = p_k \cdot (s_k - 1 ) \implies p_n = (s_k - 1)$ (since $p_n \neq p_k$) fixing $k$ and varying $k$ we see $s_k - 1$ has infinetely many divisors $\implies s_k -1 = 0 \implies s_k =1 $ since we arbirtarily chose $k$ therefore we conclude that $s_k = 1$ for all $k \in \mathbb{N} \implies a_n = p_n$ for all $n \in \mathbb{N}$ so we are done .

One liner : Notice that $p_{n}\mid{a_{n}-p_{n}}=a_{1}-p_{1}$ for all $n\ge{1}$ , thus $a_{1}=p_{1}$ and $a_{n}=p_{n}$ since the $\textbf{RHS}$ is constant . $\blacksquare$

Hey I didn't fakesolved it right? $$a_n-a_k=p_n-p_k \implies a_n-(a_k-p_k)=p_n ,p_n|a_n \implies p_n|a_k-p_k \text{For any n,k}\in\mathbb{N}$$. This implies $a_k-p_k$ has infinitely many prime factors or $a_k-p_k=0\implies a_k=p_k$ for all $k$ and we are done. Dang shouldnt have had my post wasted on this