Proof by induction $uf(n)+f(n-1)=g(n)\equiv 0\pmod x \implies f(n)\equiv 0\pmod x$

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I am trying to prove the following proposition by induction. I got stuck . Please help.

$$uf(n)+f(n-1)=g(n)\equiv 0\pmod x$$ where $u\in\mathbb Z_{\ge 0}$, and $n\in\{1,2,\cdots,p-2\}$.

If $f(0)\equiv 0\pmod x$ and $g(0)\equiv 0\pmod x$ then $f(n)\equiv 0\pmod x$.

Proof

What I plan to do:

Show that $f(1)$ is true.

Assume for some $k\in\{0,1,2,p-3\}$ that $f(k)$ is true.

Show that implies that $f(k+1)$ is necessarily true.

I got stuck from the get-go:

$uf(1)+f(0)=g(0)\equiv 0\pmod x$ $\implies$ $$uf(1)\equiv 0\pmod x$$ But how do I prove that $$f(1)\equiv 0\pmod x?$$

Any help will be greatly appreciated.Thanks