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 Integer Polynomial Illation (Posted on 2015-02-22)
Find the polynomial of the lowest degree with integer coeﬃcients such that one of its roots is:
√2 + 3√3

 No Solution Yet Submitted by K Sengupta No Rating

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 solution Comment 1 of 1
I'm going to assume that non-constant polynomials were intended, otherwise f(x)=0 is of lowest degree.

x^6-6*x^4-6*x^3+12*x^2-36*x+1

The technique I used to find this polynomial also provides the proof that it is of lowest degree as all lower degrees are eliminated in turn.

In general to find a polynomial of integer coefficients with x=2^(1/2)+3^(1/3) as a zero, I start by successively expanding
x^1
x^2
x^3
....
and in those expansions grouping terms by
2^(1/2), 3^(1/3), 3^(2/3), 2^(1/2)*3^(1/3), and 2^(1/2)*3^(2/3), and the constant term

since we are dealing with an integer combination of these powers then each of these terms must individually sum to zero.

For example,
if f(x)=a*x+b were to have the given zero then we would need
a*2^(1/2)+a*3^(1/3)+b=0
which means a=0 and a=0 and b=0 but we have eliminated the possibility of the zero polynomial.
Thus we move on to degree 2.

In a similar manner I was able to show that for degrees 2,3,4, and 5 no integer solution exists for the coefficients.

Degree 6 is the first degree to give integer solutions
namely
f(x)=x^6-6*x^4-6*x^3+12*x^2-36*x+1
and all integer multiples of this function.

 Posted by Daniel on 2015-02-23 09:20:06

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