Solution verification: Prove $f$ doesn't take the value $14$ for any integer input.

Question

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?

You could have $q(\varepsilon)=\pm 1$ without any issues, though, couldn't you? — Arthur, Jan 19 '20 at 10:47
Already answered here: Polynomials and Arithmetic: If $p(x_i) = 7$ for four integers, then $p(z)\ne14$ — TonyK, Jan 19 '20 at 10:47
@TonyK, I haven't found that thanks for contributing, but I'm still wondering if my answer is precise and correct. — PinkyWay, Jan 19 '20 at 10:51
But is $q(\varepsilon)$ required to be distinct from all the other factors? — Arthur, Jan 19 '20 at 11:53
@metamorphy, I hadn't found that until I saw the link in comments because I translated it into English how it was written in my mother tongue. Users also responded here on the question if what I wrote is correct or not, so my question is answered just like yours. — PinkyWay, Jan 19 '20 at 17:37

score 2 · Accepted Answer · edited Jan 19 '20 at 15:58

2

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 $$(n-\alpha),(n-\beta),(n-\gamma),(n-\delta)$$

edited Jan 19 '20 at 15:58

J. W. Tanner

60,406

answered Jan 19 '20 at 11:44

2

Let me fill in a step that I'd have found helpful here: Right after "let $g(x) = ...q(x)$", perhaps one should say, "Now, for contradiction, suppose that $f(n) = 14$ for some $n$; then $g(n) = 7$. But if $g(n) = 7$, then ..." – John Hughes Jan 19 '20 at 11:55
1

Yes, good idea - I've added it in the hope that it helps the OP – Jan 19 '20 at 11:58
@S. Dolan Thank you very much for the explanation and correction! – PinkyWay Jan 19 '20 at 12:12

Solution verification: Prove $f$ doesn't take the value $14$ for any integer input.

1 Answers1