I've seen a lot of questions like mine around here but most of them use the rational root theorem. However, I am not allowed to use this theorem since I haven't seen it in class. In fact, I should be able to solve this problem only by analyzing parities. Here's what I have so far:

Suppose, by contradiction, that there is at least one rational root and it is of the form $\frac{p}{q}$, $p$ and $q$ are coprimes and $q\neq0$. So we have $(\frac{p}{q})^{2018}-2(\frac{p}{q})^3+24=0$ and we can divide both sides of the equation by $q^{2018}$. Then we are left with $p^{2018}-2p^3q^{2015}+24q^{2018}=0$. Note that $2p^3q^{2015}$ and $24q^{2018}$ are always even

Case 1: $p$ and $q$ odd

This cannot be the case since odd - even + even cannot yield an even result.

Case 2: $p$ and $q$ even

$p$ and $q$ are coprime, thus this also cannot happen.

Case 3: $p$ odd and $q$ even

Similar to the 1st case, also cannot happen.

Case 4: $p$ even and $q$ odd

This is the only possible case, since even - even + even can result zero.

I should get a contradiction, but I cannot see how. Any tips? Is this the right way to prove this?

  • $\begingroup$ p even and q odd Hint: the largest power of $2$ that divides the sum of the last two terms is $2^3\,$. $\endgroup$ – dxiv Apr 11 '18 at 22:28


$p^{2018}=2(p^3q^{2015}-12q^{2018})$, so $p$ is even and then we can write $p=2k$.

$2^{2018}k^{2018}=2^4k^3q^{2015}-24q^{2018}\to 2^{2015}k^{2018}-2k^3q^{2015}=3q^{2018}$

Looking the last equation, conclude that $q$ is also even. Which is a contradiction.

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  • $\begingroup$ It should be $2(p^3q^{2015} - 12q^{2018})$ instead of $2(p^3q^{2015} + 12q^{2018})$ isn't it? $\endgroup$ – LGN Apr 11 '18 at 22:46

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