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BdMO Online Forum • View topic - Find (m , n)

Find (m , n)

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Find (m , n)

Post Number:#1  Unread postby Absur Khan Siam » Tue Jan 31, 2017 5:35 pm

Find all natural number $m,n$ such that $m^2 = 5.2^n + 1$
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
Absur Khan Siam
 
Posts: 53
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

Re: Find (m , n)

Post Number:#2  Unread postby Absur Khan Siam » Tue Jan 31, 2017 5:56 pm

My solution :
The equation can be re-written lilke this : $(m+1)(m-1) = 5.2^n$
Notice that either $(m+1)$ or $(m-1)$ can not be $5$.Because if $(m+1)$ or $(m-1)$ is $5$ then the other will also be an odd number.Thus,the L.H.S is odd.But the R.H.S. is even and so it's impossible.
So, we can say that $m+1 = 2^k$ or $m-1 = 2^k$ where $k$ is a natural number.

Case 1:
If,$m-1 = 2^k$
then , $m+1 = 2^k + 2$
Moreover , $m+1 = 5.2^q$ where $q$ is another natural number.
So,$2^k + 2 = 5.2^q \rightarrow 2(2^(k-1) + 1) = 5.2^q$
$2^(k-1) + 1$ is odd.But $5.2^q$ has only one odd factor which is $5$
Thus,$2^(k-1) + 1 = 5$ or, $k = 3,m = 9 , n = 4$.

Case 2:
If,$m+1 = 2^k$
then , $m-1 = 2^k - 2$
Moreover , $m-1 = 5.2^p$ where $p$ is another natural number.
So,$2^k - 2 = 5.2^p \rightarrow 2(2^(k-1) - 1) = 5.2^p$
Similar to the first case,$2^(k-1) - 1 = 5$
But there is no such $k$.And so,there is no solution in this case.

Therefore,we get $(m , n) = (9 , 4)$ :D
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
Absur Khan Siam
 
Posts: 53
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka


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