Bremen numbers
For some positive integers n , there exists a polynomial in n variables whose image is exactly the set of the real positive numbers (note that 0 is not included in this set). The values of n verifying that property are called Bremen numbers .
Find the sum of all the Bremen numbers smaller than or equal to 30.
Submit 0 as your answer if you believe that there does not exist any Bremen number smaller than or equal to 30.
The answer is 464.
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2 solutions
I forgot to say that 1 is not obviously a Bremen number. There are many ways to prove this.
For instance (using some topology), a 1-variable polynomial is continuous, so the image of a compact set is a compact set. When the variable goes (in absolute value) to infinity the polynomial goes (in absolute value) to infinity, so the image can not be the set of the real positive numbers. Note that this argument is not valid if the polynomial is multidimensional (just because there are "many infinities" and not 2 on a n-dimensional space).
You can also try to prove that 1 is not a Bremen number by using the "Fundamental Theorem of Algebra".
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I don't follow you in the sentence "when the variable goes (in absolute value) to infinity the polynomial goes (in absolute value) to infinity, so the image can not be the set of the real positive numbers". What if when x goes to either infinity, the polynomial goes to the same infinity (the positive one)?
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Not necessarily. It just goes to one of them (maybe to the positive or maybe to the negative one). The main point is that the polynomial does not tend to 0 when x goes to infinity (positive or negative infinity) so the image can not tend to 0. Adding the argument: "The image of a compact set it is a compact set" you conclude that the image set can not be open in 0.
Continuity of the 1-variable polynomial is not a sufficient condition, nor is it the important property. For example, e x is a continuous function whose image is R + .
You can directly work with the shape of polynomials of even and odd degree.
If n ≥ 2 , it's not hard to check that the polynomial ( x 1 x 2 − 1 ) 2 + ∑ i = 2 n x i 2 has the desired property: to get a positive real number a 2 , set x 1 = 1 / a , x 2 = a , x i = 0 for i ≥ 3 , and since it's impossible for x 1 x 2 − 1 and x 2 to be simultaneously 0 , the range of the polynomial consists of only positive real numbers.
For n = 1 , it is well-known that the image of a univariate polynomial is a closed subset of R , hence cannot be ( 0 , ∞ ) . (Indeed, univariate polynomials are "closed maps": see e.g.:
http://planetmath.org/polynomialfunctionisapropermap )
So the answer is 2 + 3 + ⋯ + 3 0 = 4 6 4 .
Well, x 2 + ( x y − 1 ) 2 is a polynomial in n variables for all n ≥ 2 , so you don't have to specify the sum ;)
At first, let's see that if n is a "Bremen number" , n+1 it is also a "Bremen number" by using induction.
If n is a "Bremen number" there exists a polynomial p(x) of n variables verifying our condition. If we call the new variable z , obviously p(x)+z·z is a polynomial of n+1 variables verifying our condition, so n+1 is a "Bremen number" .
The 2 -variable polynomial ** x·x +(1-xy)·(1-xy) verifies our conditions (obviously greater than **0 (the two addends can not be 0 at the same time), and if a is non-zero its value on (a,1/a) is a·a , which is surjective if a can be any positive (or negative) real number).
So the answer is just the addition of all natural numbers between 2 and 30 whose value is 464 (easy sum because it is an arithmetic progression)