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I am trying to show that the equation $7x^5-23x^2+2x-8=0$ has no integer solutions i.e $x$ belonging to $\mathbb{Z}$.

Now I am aware calculus can be used to show there is no real root except the rational between $1$ and $2$ but I'd like to use modular arithmetic to show the lack of integer solutions.

What I've tried so far is:

  1. Assume $k$ belonging to $\mathbb{Z}$ is a root
  2. $k(7k^4-23k+2)=8$ which shows that $8$ must divide either $x$ or $7x^4-23x+2$
  3. Take $k=8m$ and insert
  4. ?

Now this might be totally the wrong approach.

Thank you in advance.

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    $\begingroup$ You can use the rational roots theorem, can't you? According to that, all integer roots must be divisors of $8$. $\endgroup$ Commented Jan 28, 2021 at 18:32
  • $\begingroup$ @Moo corrected, thank you. $\endgroup$
    – napadia
    Commented Jan 28, 2021 at 18:36
  • $\begingroup$ Welcome to Mathematics Stack Exchange. Consider modulo $3$ $\endgroup$ Commented Jan 28, 2021 at 18:39
  • $\begingroup$ @Moo no, step 2 is correct $\endgroup$
    – napadia
    Commented Jan 28, 2021 at 18:43

3 Answers 3

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use modular arithmetic to show the lack of integer solutions:

$$7x^5-23x^2+2x-8\equiv x+x^2+2x+1\equiv x^2+1\not\equiv0\pmod3$$

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  • $\begingroup$ Thank you, how do I know to choose (mod 3)? $\endgroup$
    – napadia
    Commented Jan 28, 2021 at 18:46
  • $\begingroup$ @napadia For me, I simply tried trial and error, considering (mod n) for various small positive integers $n$. Then, I accidentally noticed that [7 + 2] = 9. $\endgroup$ Commented Jan 28, 2021 at 18:48
  • $\begingroup$ Could you explain that a bit further? $\endgroup$
    – napadia
    Commented Jan 28, 2021 at 18:52
  • $\begingroup$ @napadia See my answer. First of all, when considering a (mod 3) argument, all you have to check are $\{-1,0,+1\}.$ Clearly $1^5 \equiv 1\pmod{3}$ and $0^5 \equiv 0\pmod{3}$, so the only hard part here was recognizing that $(-1)^5 \equiv (-1)\pmod{3}.$ Then, the next challenge was recognizing that $(-23) \equiv (1)\pmod{3}$, so the problem reduced to showing that $x^2$ can't be congruent to 8 $\pmod{3}.$ $\endgroup$ Commented Jan 28, 2021 at 18:57
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    $\begingroup$ @napadia: if $7x^5-23x^2+2x-8\equiv0\pmod p$ for some $x$ between $0$ and $p-1$, then $p$ doesn't work as a modulus for this purpose $\endgroup$ Commented Jan 28, 2021 at 21:13
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$k(7k^4-23k+2)=8$

Start from here, you first would notice if $k$ is integer, then $7k^4-23k+2$ is also integer.

Therefore $k$ can be $1,2,4,8,-1,-2,-4,-8$. You should check each of these case and calculate $7k^4-23k+2$ and confirm $k(7k^4-23k+2)$ is not $8$.

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If I am not mistaken, a (mod 3) argument will work here.

$7x^5 + 2x \equiv 0 \pmod{3},~$ precisely because $x^5 \equiv x\pmod{3}.$

Then the problem reduces to noticing that $-23x^2 \equiv x^2 \pmod{3}$, which can't be congruent to $2 \pmod{3}.$

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