# Polynomials with integer coefficients - the GCD identity

#### Bernhard

I am reading Anderson and Feil on the Factorization of Polynomials - Section 5.3 Polynomials with Integer Coefficients

A&F point out that some of the therems they have proved for $$\displaystyle \mathbb{Q} [x]$$ are false if we restrict ourselves to polynomials with integer coefficients.

For example they point out that the Divsion Theorem is false for $$\displaystyle \mathbb{Z} [x]$$.

Also the GCD identity fails in $$\displaystyle \mathbb{Z} [x]$$.

They then ask the reader the polynomials 2 and x in $$\displaystyle \mathbb{Z} [x]$$ pointing out the 1 is the GCD for these polynomials. A&F then ask the reader to prove that we cannot write 1 as a linear combination of 2 and x

Can anyone help me with a rigorous and formal proof of this.

Peter

#### Deveno

MHF Hall of Honor
suppose we COULD.

we then have 1 = 2f(x) + xg(x), for some polynomials f(x),g(x) in Z[x].

suppose that deg(f) = m, and deg(g) = n. let k = max(m,n). then we can write:

$$\displaystyle f(x) = a_0 + a_1x + \cdots + a_kx^k, g(x) = b_0 + b_1x + \cdots + b_kx^k$$ (some of the coefficients may be 0).

then:

$$\displaystyle 1 = 2a_0 + (2a_1 + b_0)x + \cdots + (2a_k + b_{k-1})x^k + b_kx^{k+1}$$.

since these are equal polynomials, we must have equal constant terms, so $$\displaystyle 1 = 2a_0$$ for some integer $$\displaystyle a_0$$, which cannot happen.

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#### Bernhard

Excellent ... yes, so obvious when you see how ... Thanks so much ....

Peter