How to prove the following equality?  For $m\neq n$,  $\gcd(a^{2^m}+1,a^{2^n}+1) = 1 $ if $a$ is an even number  $\gcd(a^{2^m}+1,a^{2^n}+1) = 2 $ if $a$ is an odd number  Thanks in advance.
Prove that two distinct numbers of the form $a^{2^{n}} + 1$ and $a^{2^{m}} + 1$ are relatively prime if $a$ is even and have $\gcd=2$ if $a$ is odd.   My attempt: If $a$ is even, let $a = 2^{s}k$ for some integers $k, s$ Then, $$a^{2^{n}} + 1 = 2^{2^{n}s}\cdot k^{2^n} + 1$$ and $$a^{2^{m}}  + 1 = 2^{2^{m}s}\cdot k^{2^m} + 1$$ To prove that they're relatively prime, we need to show that their gcd = 1. And I was stuck here, how could I prove that gcd of two numbers is $1$?     A hint would be sufficient. Thanks.
Plotting the function $f(x)=x^{1/3}$ defined for any real number $x$ gives us:   Since $f$ is a function, for any given $x$ value it maps to a single y value (and not more than one $y$ value, because that would mean it's not a function as it fails the vertical line test). This function also has a vertical tangent at $x=0$.   My question is: how can we have a function that also has a vertical tangent? To get a vertical tangent we need 2 vertical points, which means that we are not working with a "proper" function as it has multiple y values mapping to a single $x$. How is it possible for a "proper" function to have a vertical tangent?  As I understand, in the graph I pasted we cannot take the derivative of x=0 because the slope is vertical, hence we cannot see the instantaneous rate of change of x to y as the y value is not a value (or many values, which ever way you want to look at it). How is it possible to have a perfectly vertical slope on a function? 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