Let $f\in\mathbb Z[x]$ be a polynomial with integer coefficients, s.t. it takes the value $7$ for $4\;\text{distinct integers}$. Prove $f$ doesn't take the value $14$ for any integer input.
My attempt: $$f(\alpha)=f(\beta)=f(\gamma)=f(\delta)=7,\;\;\alpha\ne\beta\ne\gamma\ne\delta\in\mathbb Z$$ From the fact: $$\forall x,y\in\mathbb Z\;\;x\pm y\in \mathbb Z$$ $$g(x):=f(x)-f(\alpha)=f(x)-f(\beta)=f(x)-f(\gamma)=f(x)-f(\delta)$$ $$\implies g(x)=(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)q(x)$$ Let $\varepsilon\in\mathbb Z$. $$f(\varepsilon)=14$$ By the above definition of $g(x)$, for $g(\varepsilon)$: $$g(\varepsilon)=(\varepsilon-\alpha)(\varepsilon-\beta)(\varepsilon-\gamma)(\varepsilon-\delta)q(\varepsilon)$$ $$\alpha\ne\beta\ne\gamma\ne\delta\ne\varepsilon\in\mathbb Z\implies\;(\varepsilon-\alpha)\ne(\varepsilon-\beta)\ne(\varepsilon-\gamma)\ne(\varepsilon-\delta)\in\mathbb Z$$ $14$ can be factorised by at most $4\;\text{distinct integers}$,e.g.: $$14=1\cdot(-1)\cdot(-2)\cdot7\leftarrow\;\text{one permutation}$$ Therefore,$f$ doesn't take the value $14$ for any integer input. Is this legitimate?
The idea of your argument is excellent, and it can be written correctly as follows.
Let $g(x)=f(x)-7$; then $$g(x)=(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)q(x).$$
Now, for contradiction, suppose that $f(n)=14$ for some $n$; then $g(n)=7$. But if $g(n)=7$ for some integer $n$, then $$7=(n-\alpha)(n-\beta)(n-\gamma)(n-\delta)q(n).$$ $7$ can be expressed as the product of at most 3 distinct integers and so cannot have the four distinct factors $$(n-\alpha),(n-\beta),(n-\gamma),(n-\delta)$$