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anonymous · Answer

Find all integers n such that $$4^{(n-1)/(n+1)}$$ is integer

kinggeorge · Answer

That answer wasn't actually correct give me a minute.

anonymous · Answer

sure

kinggeorge · Answer

I got a little mixed up. It's actually equivalent to finding all $n$ such that $${n-1 \over n+1}$$is a multiple of $1\over2$. In other words, find all $n$ such that $${n-1 \over n+1}={k\over2} $$for some $k\in\mathbb{Z}$

anonymous · Answer

what after that?

kinggeorge · Answer

Cross multiply, and get all the $n$'s on one side. You should get $$kn-2n=-k-2=n(k-2)$$

anonymous · Answer

but the ans contains finite nos. in a set

kinggeorge · Answer

I suppose I'm trying to do it the hard way. An easier way to do it would be to let $n+1$ be equal to $2, 4, 6$. If it's equal to 2, then $n=1$, so your exponent would be $$4^{{0\over1}}=4^{0}=1$$Also, if $n+1=4$, then $n=3$ and $$4^{2\over4}=4^{1\over2}=2$$Finally, if $n+1=6$, then $n=5$ so $$4^{4\over6}=4^{2\over3} \notin \mathbb{Z}$$Thus, if we continue the pattern, as n gets larger, the exponent reduces to something not of the proper form. So if I'm finally correct, the answer should be $n=2, 4$

anonymous · Answer

the ans has negative nos too

kinggeorge · Answer

Those should be $n= -1, -2$ by the same reasoning as above.

Also, above, $n=1, 3$ not $2, 4$ since it's $n+1=2, 4$.