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#1 h Aug 18, 2023, 3:24 PM

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Let $f(x)$ be a polynomial with integer coefficients such that $f(0)=2020$ and $f(a)=2021$ for some integer $a$ . Prove that there exists no integer $b$ such that $f(b) = 2022$ .

This post has been edited 4 times. Last edited by Entrepreneur, Aug 18, 2023, 3:30 PM

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#2 h May 11, 2025, 10:32 AM

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Upon inspection, we see that $2022 = 2(2021)-2020$ . Let $f(x) = \sum_{k=0}^{n}c_{k}x^{k}$ such that $f(0)=2020, f(a)=2021, f(b)=2022 \Rightarrow f(b)=2f(a)-f(0)$ for $a,b,c_{k} \in \mathbb{Z}$ . This yields:

$f(b) = 2\sum_{k=0}^{n} c_{k}a^{k} - \sum_{k=0}^{n}c_{k}0^{k} = \sum_{k=0}^{n}c_{k}(2a^{k}-0) = \sum_{k=0}^{n}c_{k}b^{k} \Rightarrow b^{k}=2a^{k} \Rightarrow \frac{b}{a} = 2^{1/k}$

which is a contradiction since a rational number $\neq$ an irrational number. Thus, no such $b \in \mathbb{Z}$ exists under these conditions.

This post has been edited 4 times. Last edited by Mathzeus1024, May 11, 2025, 10:42 AM

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#3 h May 11, 2025, 11:04 AM

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This is not true, take $f(x)=x+2020$ and $a=1,b=2$ .

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#4 h May 11, 2025, 11:42 PM

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Mathzeus1024 wrote:

$\sum_{k=0}^{n}c_{k}(2a^{k}-0) = \sum_{k=0}^{n}c_{k}b^{k} \Rightarrow b^{k}=2a^{k} \Rightarrow \frac{b}{a} = 2^{1/k}$

which is a contradiction since a rational number $\neq$ an irrational number. Thus, no such $b \in \mathbb{Z}$ exists under these conditions.

You can't assume that $b^k=2a^k$ just because $\sum_{k=0}^{n}c_{k}(2a^{k}) = \sum_{k=0}^{n}c_{k}b^{k}.$ In fact, this clearly wrong when $k=0$ and $2^{\frac 1k}$ isn't irrational when $k=1$ , yet such $f,a,$ and $b$ exist as alexheinis pointed out. The rest of your solution is neat, its just this one step that makes an assumption. In fact, I believe the statement is false for all $f$ except constant and quadratic polynomials. Here is a proof:

The constant case is clear and the construction above proves the linear case. If $f$ has degree $n\geq 3$ , then take
$f(x)=g(x)x(x-1)(x-2)+x+2020,$ where $g$ has degree $n-3$ with integer coefficients. Then $f(0)=2020, f(1)=2021,$ and $f(2)=2022.$ (This type of construction is common when you have one polynomial behaving like a smaller degree polynomial, such as $x+2020$ , for a set of $x$ .) For the quadratic case, take $f(x)=cx^2+dx+2020$ . Then $f(a)=2021$ and $f(b)=2022$ give, respectively,
$ca^2+da=a(ca+d)=1$ and
$cb^2+db=b(cb+d)=2.$ The first equation tells us that $a=\pm 1,$ from which we get
$a=1\implies c+d=1$ and
$a=-1\implies -c+d=-1.$ We can do similar analysis for the second equation:
$b=2\implies 2c+d=1,$ $b=1\implies c+d=2,$ $b=-1\implies -c+d=-2,$ and
$b=-2\implies -2c+d=-1.$
It is easy to see that no pair of $a$ and $b$ produce a nonzero, integral value for $c$ (nonzero since we are looking for quadratics), and hence no such quadratics exist.

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#5 h May 12, 2025, 12:25 AM

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We have that $\frac{P(a)-P(b)}{a-b}$ is an integer for integers $a$ and $b$
so we have

$\frac{2021-2020}{a-0}$ so $a=1$ or $a=-1$

We also have $\frac{2022-2020}{b}$ so $b=-2,-1,1,2$

Lastly we have $\frac{2022-2021}{b-a}$ so $b-a=-1,1$

and then after doing all that we realize the whole problem is wrong because $f(x)=2020+x$ works.

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#6 h May 12, 2025, 12:33 AM

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I don't understand any of the math above, but this seems fairly simple like $x+2021=y$ where $a=0, b=1$

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