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 :(

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- Jan 29th 2013, 12:48 PMdarenceequation and her solutions...
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 :( - Jan 29th 2013, 02:29 PMDevenoRe: equation and her solutions...
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. - Jan 29th 2013, 02:36 PMemakarovRe: equation and her solutions...
Yes, but the question is about $\displaystyle a^n + b^n$ and not $\displaystyle (a+b)^n$.

- Jan 29th 2013, 02:53 PMDevenoRe: equation and her solutions...
- Jan 29th 2013, 03:42 PMdarenceRe: equation and her solutions...
thank you very much.

How to proove a^n+b^n is not divisible with 5? - Jan 29th 2013, 03:43 PMemakarovRe: equation and her solutions...
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.