equation and her solutions...

If a and b are solutions of equation , proove that number is integer not divisible with 5.

Please help, no idea :(

since x² - 6x + 1 = (x - a)(x - b) = x² - (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 aⁿ + bⁿ 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 aⁿ + bⁿ = (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.

Yes, but the question is about and not .

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Originally Posted by emakarov

Yes, but the question is about and not .

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its funny how your mind plays tricks on you. i noticed this just as i posted my answer, and amended it accordingly.

thank you very much.
How to proove a^n+b^n is not divisible with 5?

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Originally Posted by Deveno

aⁿ + bⁿ = (a^n-1 + b^n-1)(a + b) - (ab)(a^k-2 + b^k-2)

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And concerning divisibility by 5, it is sufficient to consider this recurrence relation modulo 5: this is a periodic sequence.

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Originally Posted by Deveno

its funny how your mind plays tricks on you.

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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.