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2^n=n^2

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by oldgeezer » Tue Jul 29, 2008 7:30 pm
for 2? (2 and 4)
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by parallel_chase » Wed Jul 30, 2008 1:27 am
I'd agree with oldgeezer, only 2 and 4 are the possible values of N in such expression.
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by acmeheadshot » Thu Jul 31, 2008 6:24 am
n 2^n n^2
0 1 0
1 2 1
2 4 4
3 8 9
4 16 16
5 32 25
6 64 36
.
.
.
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by VP_RedSoxFan » Thu Jul 31, 2008 7:15 am
I like the approach by acmeheadshot. If you run a quick chart of possible N and then the solutions to 2^N and N^2 you can

a) see that there are 2 (N=2,4) solutions for N,

b) that as N gets bigger 2^N and N^2 move farther apart so we can be reasonable sure there aren't more as N gets bigger, and

c) (this is not shown in acmeheadshot's post but I'm sure he/she thought about it anyway) if we throw a few negative integers for N, because the problem doesn't preclude them, we can know that no negative integers are solutions for N
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