Do there exist $a(x),b(x)\in\Bbb{Q}[x]$ s.t. $(x^4+4x^3-7x+2)a(x)+(x^2+3x-4)b(x)=x^2+1$?

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Prove or disprove the following:

There exist $a(x), b(x) \in \Bbb Q[x]$ such that $a(x)f(x) + b(x)g(x)=x^2+1$, where $f(x)=x^4+4x^3-7x+2$ , $g(x)=x^2+3x-4$.

It looks very much like Bezout's identity, although $\gcd(f,g)\ne x^2+1$, so I can't use it. ($\gcd(f,g)=x-1$)

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You have done most of the work

Consider $\gcd(f,g)=x-1$

This means that $f(x)=(x-1)(F(x))$ and similarly $g(x)=(x-1)(G(x))$.

We can substitute this into the equation

$$(x-1)(a(x)F(x)+b(x)G(x))=x^2+1$$

But $x-1$ does not divide $x^2+1$