# Show that the equation has no integer solutions

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.

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

use modular arithmetic to show the lack of integer solutions:

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

• Thank you, how do I know to choose (mod 3)? Commented Jan 28, 2021 at 18:46
• @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. Commented Jan 28, 2021 at 18:48
• Could you explain that a bit further? Commented Jan 28, 2021 at 18:52
• @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 $\pmod{3}.$ Commented Jan 28, 2021 at 18:57
• @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 Commented Jan 28, 2021 at 21:13

$$k(7k^4-23k+2)=8$$

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 $$8$$.

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 $$2 \pmod{3}.$$