\(n\)-এর সকল সম্ভাব্য মানের যোগফল বের করো যাতে \(n\), \(n^2+10\), \(n^2-2\), \(n^2-8\), \(n^3+6\)-এর সবগুলো মৌলিক সংখ্যা হয়। (হিন্ট: এরকম অন্তত একটি \(n\) রয়েছে)।
Find the sum of all possible \(n\) such that \(n\), \(n^2+10\), \(n^2-2\), \(n^2-8,\) \(n^3+6\) are all prime numbers. (Hint: there is at least one such \(n\).)
BdMO National Junior 2020 P11
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- Enthurelxyz
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Re: BdMO National Junior 2020 P11
Let, $n\neq 7$
So, $n \equiv 1,2,3,4,5,6 (mod$ $7)$ $
$n^2 \equiv 1,2,4 (mod$ $7)$.
If $n^2 \equiv 1 (mod$ $7)$ then $n ^2-8\equiv 0 (mod$ $7)$. So, $n^2=115$ but $15$ is not a perfect square.
If $n^2\equiv 2 (mod$ $7)$ then $n^2-2=7
n=3$ $but $3^2-8=1$ which is not a prime.
If $n^2 \equiv 4 (mod$ $7)$ then $n^2+10\equiv 0(mod$ $7)$
$ n^2+10=7$ but it has no real solution.
If $n=7$ then $n^2+10,n^2-2,n^2-8,n^3+6$ are all prime.
So, $n=7$ and answer is $7$.
So, $n \equiv 1,2,3,4,5,6 (mod$ $7)$ $
![Arrow :arrow:](./images/smilies/icon_arrow.gif)
If $n^2 \equiv 1 (mod$ $7)$ then $n ^2-8\equiv 0 (mod$ $7)$. So, $n^2=115$ but $15$ is not a perfect square.
If $n^2\equiv 2 (mod$ $7)$ then $n^2-2=7
![Arrow :arrow:](./images/smilies/icon_arrow.gif)
If $n^2 \equiv 4 (mod$ $7)$ then $n^2+10\equiv 0(mod$ $7)$
![Arrow :arrow:](./images/smilies/icon_arrow.gif)
If $n=7$ then $n^2+10,n^2-2,n^2-8,n^3+6$ are all prime.
So, $n=7$ and answer is $7$.
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