BdMO National Higher Secondary 2008/7

[Moon]

Problem 7:
We want to find all integer solutions $(m, n)$ to \[1+ 5\cdot 2^m = n^2\] .
(A) Find an expression for $n^2 -1$ ;
(B) are $(n +1)$ and $(n -1)$ both even, or both odd, or is one even and the other odd?
(C) Let $a=\frac {n-1}{2}$, Find an expression for $a(a +1)$
(D) If $a$ is odd, is $a +1$ even or odd?
(E) From parts (C) and (D), is it possible for $a = 1$, or $a(a +1) = ?$
(F) Find the only possible values $a$ can take and then find what $m $ and $n$ should be.

[samiul_samin]

(a) $n^2-1=5. 2^m$

(b) both $(n+1)$ and $(n-1)$ are even.As $n$ is always odd.

(c)$((n^2-1)/4)$

(d)$a+1$ is odd.

(e)No, it is not possible.

(f) only possible value of $a$ is $4$ and $m=4$ and $n=9$