Solution of $ 1+ 5 * 2^m =n^2$ Equation and some other question from this equation I want to find all integer Solutions of $ 1+ 5 * 2^m =n^2$ Equation . From this eqn I have to answer the following answer .

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*I have to find an expression for $ n^2-1 $


*Are $ (n+1) $ and $ (n-1) $ both even or both odd or is one even and the other odd ?


*Let $ a = \frac{n-1}{2} $ . I have to find an expression for $ a(a+1) $ .


*If a is odd , is $ (a+1) $ even or odd ?


*From parts 3 and 4 , is it possible for $ a(a+1) =1 $ or $ a=1 $ ?


*I have to find the pnly possible values a can take and then have to find what m and n should be ?
I want to have the solution of the above equation . Then I will answer the above question .

I am considering the case $m\ge 2$
Observe that $5\cdot2^m=n^2-1$ is even and so are $n\pm1$ 
and $(n-1,n+1)=(n-1,n+1-(n-1))=(n+1,2)=2$
alternatively, if integer $d>0$ divides both $n+1,n-1, d$ will divide $n+1-(n-1)=2$ 
$\implies (n-1,n+1)|2\implies (n-1,n+1)=2$ as $n\pm1$  are even
So, we have
$$5\cdot2^{m-2}=\frac{n+1}2\cdot \frac{n-1}2$$ with  $\left(\frac{n+1}2,\frac{n-1}2\right)=1$
So, either $\displaystyle\frac{n+1}2=5; \frac{n-1}2=2^{m-2}$
or $\displaystyle\frac{n-1}2=5; \frac{n+1}2=2^{m-2}$
or $\displaystyle\frac{n-1}2=5\cdot2^{m-2} ;\frac{n+1}2=1$
or $\displaystyle\frac{n+1}2=5\cdot2^{m-2} ;\frac{n-1}2=1$
