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Let the polynomial $f(x)=x^n+a_1 x^{n-1}+a_2 x^{n-2}+.....a_{n-1}x+a_n$ have integral coefficients. If there exists four distinct integers $a,b,c$ and $d$ such that $f(a)=f(b)=f(c)=f(d)=5$ show that there is no integer $k$ such that $f(k)=8$

I have tried to prove it but I somehow feel that my proof is incorrect please point out the errors and suggest some other way to do the question. This how I proceeded $\frac{x^{n+1}-1}{x-1}=x^n+a_1 x^{n-1}+a_2 x^{n-2}+.....a_{n-1}x+a_n $

$\frac{a^{n+1}-1}{a-1}=5$ hence $\frac{a^{n+1}-1}{5}=a-1$ Since $a$ is an integer $a-1$ will also be an integer hence $5|a^{n+1}-1$

Using fermat's little theorem we know that $a^4-1\equiv_5 0$ Since $f(a)=5$ we get $n+1=4$

Let k be an integer such that $f(k)=8$ $\frac{a^4-1}{8}=a-1$ Since $a-1$ is an integer $8|a^4-1$

We will consider two cases

  1. when $a$ is an even integer
  2. when $ a$ is an odd integer

If $a$ is even then $a^4-1$ will be odd and it not be divisible by $8$

If $a$ is odd then it will be relatively prime to $8$ hence $8$ will be Carmichael number

$a^7-1\equiv0(mod8)$ which proves that $a^4-1$ is not divisible by 8. $\therefore$ there is no integer k such that $f(k)=8$

Pulsar
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2 Answers2

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Hint: show that if $f(a) = f(b) = f(c) = f(d) = 5$, then

$$f(x) = 5 + (x-a)(x-b)(x-c)(x-d) g(x)$$

for some polynomial $g$ with integer coefficients. If $f(e) = 8$, then

$$(e-a)(e-b)(e-c)(e-d) g(e) = 3,$$

but a product of four distinct integers can't be a divisor of $3$.

Adayah
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  • This is exactly what I would have done, +1. However, a question for @Pulsar. Do you know how to prove that $g$ is in $\mathbb Z[x]$? This is essential for this proof to work. – Ennar Aug 27 '17 at 11:33
  • @Ennar I could not prove that can you please help? – Pulsar Aug 27 '17 at 12:34
  • @Pulsar, it is too long for a comment. The sketch would be something like this: let $f, g\in \mathbb Z[x]$ such that $g$ is primitive and $q\in\mathbb Q[x]$ such that $f = qg$. Then $q\in\mathbb Z[x]$. The proof relies on Gauss's lemma, specifically that product of two primitive polynomials is primitive. Another way is to directly apply the fact that both $\mathbb Z[x]$ and $\mathbb Q[x]$ are unique factorization domains. – Ennar Aug 27 '17 at 13:11
  • @Ennar thanks a lot those articles are very useful – Pulsar Aug 27 '17 at 13:23
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    @Ennar Are you sure this is necessary? I thought it was a lot easier: if $R$ is any ring, we can divide by monic polynomials staying within $R[X]$. If $R$ is also an integral domain, the process of pulling out the factors $x-x_i$ can be continued, so we get the factorization from the answer. – Adayah Aug 27 '17 at 14:00
  • Adayah, ok, you are right, we can do that. Not only for monic polynomials, but leading coefficient can be any unit. I just immediately skipped to $\mathbb Q[x]$ being Euclidean domain, while the proof of $\mathbb F[x]$ being Euclidean, where $\mathbb F$ is a field, uses this lemma. – Ennar Aug 27 '17 at 14:10
  • @Ennar Sorry, which lemma does this fact use? I thought the proof that $\deg : \mathbb{F}[x] \setminus { 0 } \to \mathbb{N}$ is an Euclidean function is a straightforward induction. :p – Adayah Aug 27 '17 at 14:23
  • Well the lemma you just mentioned, if $f,g\in R[x]$ and $g$'s leading coefficient is unit in $R$, then there exists $q, r\in R[x]$ such that $f = qg + r$. You can incorporate this directly for $R$ field and do what you say. – Ennar Aug 27 '17 at 14:28
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I think the first part of your solution is wrong ($\frac{x^{n+1}-1}{x-1}=x^n+x^{n-1}+\cdots+x+1$, not $x_n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$).

Since $f$ is a polynomial with integer coefficients, we have $a-b|f(a)-f(b)$ for all $a, b\in \mathbb{Z}, a\neq b$. Thus if we suppose there exists integer $k$ with $f(k)=8$, we have $k-x|f(k)-f(x)=3$, where $x=a,b,c,d$. Since $a,b,c,d$ are all distinct we have $\{a,b,c,d\}=\{k-3,k-1,k+1,k+3\}$.

Then, consider $g(x)=f(x-k)-5$. $g$ clearly is a monic polynomial with integer coefficients such that $g(0)=3, g(-3)=g(-1)=g(1)=g(3)=0$. If we set $g(x)=x^n+b_1x^{n-1}+\cdots+b_{n-1}x+b_n$, by conditions we have $b_n=3$ and following:

$3^n+3^{n-1}b_1+\cdots+3b_{n-1}+3=0 \rightarrow 3^{n-1}+3^{n-2}b_1+\cdots+b_{n-1}+1=0$

$(-3)^n+(-3)^{n-1}b_1+\cdots+(-3)b_{n-1}+3=0 \rightarrow (-3)^{n-1}+(-3)^{n-2}b_1+\cdots+b_{n-1}-1=0$

From the first equation we have $b_{n-1}\equiv-1 \mod 3$, while from the second equation we have $b_{n-1}\equiv 1 \mod 3$, a contradiction. Thus there exists no such $f$.

shrimpabcdefg
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