Prove that there is no complex root such that both the Re and Im parts are rational.

mathwizard · #1 $Old$ June 18th, 2008, 06:42 PM

Let $P$ be a polynomial with integer (real) coefficients such that; (1) - the coefficient of the leading term and the coefficient of the independent term (counted as the coefficient of $x^0, 0$ degree) are odd; (2) -The total number of odd coefficients is odd.

Like in
$P(x) = x^3 - 5x^2 + 2x -7$
$P(x) = 9x^3 - 6 x^2 + 3x -5$
$P(x) = x^4 +5x^3 + 7x^2 + x +1$
$P(x) = 7x^5 + 2x^4 - x^3 + 2x^2 - 8x -3$
Prove that $P$ has no root such that both the real and the imaginary parts are rational. In other words, if $a + bi$ is a root of $P$ , then at least one of the numbers $a$ and $b$ is irrational

Opalg · #2 $Old$ June 20th, 2008, 03:14 PM

Let $P(x) = a_nx^n + \ldots+a_1x+a_0$ , and suppose that $x = \textstyle\frac{p+iq}r$ is a root, where p, q and r are integers with no common factor. Then

. . . . . $a_n(p+iq)^n + a_{n-1}(p+iq)^{n-1}r + \ldots + a_1(p+iq)r^{n-1} + a_0r^n = 0$ .. . . . . (*)

If r is even then 2=(1+i)(1-i) divides $a_n(p+iq)^n$ . Since 1±i are Gaussian primes which do not divide a_0, they must divide p+iq. Thus p and q are both even, contradicting the assumption that p, q and r have no common factor. Therefore r is odd.

If 1+i divides p+iq then it divides r (since it cannot divide the odd number a_0). But then 1-i will also divide r, and so r is even. This is another contrdiction, and we conclude that 1+i does not divide p+iq.

Next, notice that 1+i divides the Gaussian integer m+in if and only if m and n have the same parity (both even or both odd). If we call a Gaussian integer "even" if it is a multiple of 1+i, and "odd" if it is not, then "even" and "odd" have the same additivity and multiplicativity properties as they do for real integers. That is, the sum of two "odd" Gaussian integers is "even", as is the sum of two "even" integers, whereas the sum of an "odd" integer and an "even" integer is "odd". Also, the product of two Gaussian integers is "even" unless both of them are "odd", in which case the product is "odd".

Coming back to (*), you see that the left side contains an odd number of "odd" terms, and is therefore "odd". But the right side is 0, which is "even". That is the final contradiction, showing that the original assumption (that (*) has a solution with rational real and imaginary parts) is false.