If a and b are solutions of equation $\displaystyle x^2-6x+1=0$, proove that $\displaystyle \forall n \in N$ number $\displaystyle a^n+b^n$ is integer not divisible with 5.
Please help, no idea
since x^{2} - 6x + 1 = (x - a)(x - b) = x^{2} - (a+b)x + ab,
we have a + b = 6, which is clearly an integer. note 5 does not divide 6. note also that ab = 1, as we will need this later.
also (a + b)^{2} = (a + b)(a + b) - 2ab = 36 - 2 = 34, which is also an integer.
now let's prove a^{n} + b^{n} is an integer, by induction on n.
suppose a^{k-1} + b^{k-1} is an integer, for all k < n as our induction hypothesis.
then a^{n} + b^{n} = (a^{n-1} + b^{n-1})(a + b) - (ab)(a^{k-2} + b^{k-2})
since the RHS side is all integers, the LHS must be, as well.
And concerning divisibility by 5, it is sufficient to consider this recurrence relation modulo 5: this is a periodic sequence.
I can relate. On the other forum, there was a discussion about formal proofs. I argued that verifying proofs on the computer ensures their correctness for all practical purposes, but agreed that there is no absolutely 100% guarantee, if only for this reason. Though typos can usually be eliminated by a community of researchers.