There is one other possibility, , so I take your point that requiring isn't really necessary... but it does open the opportunity to expand the proof question to ask for all possible solutions if M is an integer, including a proof that there are no other possibilities with .Is the condition really necessary?
The only case of this I can see is , so it would be simpler to leave it unmentioned.
You could go into complex numbers and use Taylor series expressions for real and imaginary part, then you could prove local finite integer ABSOLUTE convergence from a point, calling z_n = M^n + M^(-n) and by proving that |Re(z_n)| and |Im(z_n)| ==> |z_n| converges. Type of approach from complex analysis.Prove that if but , then for all .
I am curious as to how Extension 2 students would approach this as a proof problem, and wondering if there are other approaches beyond the two that occur to me.
Yes, it will work, but it is effectively a strong induction proof:Does my Binomial expansion not work, cus that and Induction is the only way My small brain can think off. was thinking of complex numbers too since its in that form but yeah haha
This is completely valid but technically requires induction to complete the argument for all naturals.I was thinking about a proof like this: The quadratic equation with roots at is
, which expands and rearranges to give
It follows that
We then prove that , with or without using induction,
because that gives must be an integer.
The logic is inductive, I agree, though it doesn't actually need to be presented as an induction proof.This is completely valid but technically requires induction to complete the argument for all naturals.