Should be a simple proof

[carmine current]

I have an idea but i dont know how to do this more formally , point out if i missed something

Let a=x and b=1/x such that a+b is an integer.
If a is an integer , then a^n is an integer too.
If a is a Real number then , the choice of an appropriate b will suffice to make a+b an integer, which inturn makes a^n + b^n an integer.
If b is an integer, then b^n is an integer too.
If b is a real number , then b should be choosen appropriately so that a+b is an integer.
Hence a^n + b^n is an integer too , as the sum of two integers is an integer.

[lean swift]

try expanding (x + 1/x)^n

uh actually i don't know if this will work

yeah nevermind, sorry