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Let $a_n=2^n+6^n+3^n-1^n$,find all $n$ such that $a_n$ is coprime to all other elements of this sequence.

Probably the problem statement should be:

Determine all positive integers such that they are coprime to all the terms of this sequence
\[a_n=2^n+6^n+3^n-1^n, \forall n \in N\]

আর যে এই সমস্যার সমাধান একবার দেখছে তার ভোলার কথা না আমি খালি সমাধান মনে করে লিখতেছি না, কেউ যদি এখন সমাধান করে তাহলে সে সমাধান দিক

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শুভ জন্মদিন, মাসুম

Thanks.Now I am 17 .And this is the easiest problem of Imo(according to me).Just note that $a_n$ is even for all $n$.So $gcd(a_m,a_n)$ is at least $2$,therefore never coprime.

Masum's solution is correct if the problem is as stated by himself but if the problem is as stated by Zzzz (:P) then it's incorrect.

But Zzzz,what do you mean by 'determine all positive integers'-$n$ or $a_n$

Zzzz wrote:Probably the problem statement should be:

Determine all positive integers such that they are coprime to all the terms of this sequence
\[a_n=2^n+6^n+3^n-1^n, \forall n \in N\]

আর যে এই সমস্যার সমাধান একবার দেখছে তার ভোলার কথা না আমি খালি সমাধান মনে করে লিখতেছি না, কেউ যদি এখন সমাধান করে তাহলে সে সমাধান দিক

TIURMI,I didn't undestand your comment too.

Because never is $gcd(a_m,a_n)=1,$ I think there exists neither $n$ nor $a_n$ satisfying the condition.

\[a_1=6^1+2^1+3^1-1=10\]
\[a_2=6^2+2^2+3^2-1=48\]
\[a_3=6^3+2^3+3^3-1=250\]
\[.\]
\[.\]
\[.\]

So the sequence is $10,48, 250,...,\infty$

In the IMO question, they asked for all positive numbers which are coprime to all numbers of the sequence. For example, $1$ is such a number. $1$ is coprime to all the numbers of the sequence. Now look for others !

Well Masum your mistake is you thought that you are to find number "in the sequence" which are coprime to the other numbers of the sequence but the question states that you have to find "all" the positive integers which are "within" or "out of the sequence" that are coprime with any $a_n$ (i.e. all of the numbers in the sequence)