\(m\) আর \(n\) হচ্ছে এমন ধনাত্মক পূর্ণসংখ্যা যাতে \(1 + 2^m = n^2\) হয়। \(10m+n\)-এর সকল সম্ভাব্য মান এর যোগফল বের করো।
\(m\) and \(n\) are positive integers such that \(1 + 2^m = n^2\). Find the sum of all possible values of \(10m+n\).
BdMO National Junior 2020 P1
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Anindya Biswas
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Re: BdMO National Junior 2020 P1
It is easy to check that $m=3,n=3$ gives us the smallest solution.
Let's manipulate this equation like this:
$2^m=(n-1)(n+1)$
So, only prime divisor of $n-1$ and $n+1$ is $2$
So, we can assume that $n-1=2^a$, $n+1=2^b$
It is easy to check that $1\leq a<b$ since $a=0$ doesn't give any solution.
And $2^b-2^a=2\Rightarrow 2^{b-1}=1+2^{a-1}$
The left side is even. The right hand side would be even only if $a-1=0\Rightarrow a=1$.
This gives us $b-1=1\Rightarrow b=2$ which gives the only solution $m=3, n=3$.
So, our answer is $33$
Let's manipulate this equation like this:
$2^m=(n-1)(n+1)$
So, only prime divisor of $n-1$ and $n+1$ is $2$
So, we can assume that $n-1=2^a$, $n+1=2^b$
It is easy to check that $1\leq a<b$ since $a=0$ doesn't give any solution.
And $2^b-2^a=2\Rightarrow 2^{b-1}=1+2^{a-1}$
The left side is even. The right hand side would be even only if $a-1=0\Rightarrow a=1$.
This gives us $b-1=1\Rightarrow b=2$ which gives the only solution $m=3, n=3$.
So, our answer is $33$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann