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The sequence is defined by and
for all positive integers . Determine all prime numbers for
which there exists a positiveinteger such that divides the number
.
Define . Easy to see that . . Obvious
conclusion is that each is fo the form . Now we can check
that
, where is Fibonacci's sequence.
We will prove that every prime satisfies the condition. Let (for
it's
trivial). . Using Fermat's Little Theorem
we know that if and we will be done. But we know that if
a pair of 2 consecutive Fibonacci's numbers is mod equal to , it
determines
next and previous pair mod , so if we consider pairs mod as
vertices and put
edge between and , it forms cycle covering. We have to prove
that pair
lies on the same cycle as . But this is true, because there's
edge from
to and edge from to .
The sequence is defined as follows: .Prove that all the terms of
the sequence are integers.
We proceed with induction. , so is an integer (as are and ).
Now, assume
that is an integer for . We have that . Upon
multiplying out, we have that . Now, and
are coprime since . Thus, it suffices to show that
(which is clearly true since ) and
(which is clearly true since ). Since
both are proven, is an integer, so induction is complete.
Let be a given real number and assume that the polynomial
satisfies
, for . Prove that the degree of is at least .
uppose is a polynomial of degree less than . The polynomial
is the unique polynomial of degree less than or equal to given
the values of for. So that polynomial must be and the term must be
0. Therefore,
But we have for all , so that
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a contradiction. So must have degree at least .
Suppose . Observe that each of the products have
degree , so has degree at most . Also, because the product in
the 0th
term in the sum is and the product in the th term for is
.
In a similar way, . If
, then has degree at most but roots, so must be the zero
polynomial, giving
. Of course, usually will have degree , so the condition that
has degree
strictly less than requires the coefficient of to be 0.
Define the polynomial for and .
Then
has degree and leading coefficient for all ,
and for all we have
.
So and for all .
scorpius119's argument shows, that this is in fact the only
polynomial of degree
with for all and some .
Let be a real quadratic trinomial, so that for all the
inequality
holds. Find the sum of the roots of .
Let . Then
. Since we needfor all , we need , i.e. . Therefore
, and so by Vite's relation, the sum of the roots of is
.
Indeed for all (since the discriminant of
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