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Let $a, b$ be two distinct positive integers such that both $\sqrt[4]{a}$ and $\sqrt[4]{b}$ are irrational. Can $\sqrt[4]{a} - \sqrt[4]{b}$ be a rational number? Should that be the case, I would be quite surprised!

NOTE. The analogous question with the fourth root replaced everywhere by the square root is easily seen to have a negative answer: see Iulu's answer to the post Integer Difference Between Roots. More generally Edge Erdil states in his answer to the post Is the difference of two irrationals which are each contained under a single square root irrational that the answer is negative if we replace the fourth root everywhere by the p-th root, with p prime.

Maurizio Barbato
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  • This difference cannot be rational. But how familiar you are with concepts from the theory of field extensions (up to and possibly including Galois theory)? That affects what kind of arguments you will understand. – Jyrki Lahtonen Nov 23 '22 at 13:24
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    If you know about algebraic integers, then it is immediate that the difference can be rational only if it is an integer, and that opens up other possibilities. So, what tools / pieces of theory are you familiar with? – Jyrki Lahtonen Nov 23 '22 at 13:39
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    Barring a calculation error the number $s=\root4\of{a}-\root4\of{b}$ is a zero of the polynomial $$x^{16} -4(a+b)x^{12}+ \left(6 a^2-124 a b+6 b^2\right)x^8+ \left(-4 a^3-124 a^2 b-124 a b^2-4 b^3\right)x^4+(a-b)^4.$$ So the rational root test tells us that if $s$ is rational, it must be an integer and a factor of $(a-b)^4$. – Jyrki Lahtonen Nov 23 '22 at 13:48
  • @JyrkiLahtonen Dear Jyrki, I have an elementary knowledge of Galois theory. I have briefly touched topics as field extensions, Galois groups, Galois correspondence and the fundamental theorem of Galois theory. But I have little, if nothing, practical understanding of these subjects, since I have never or almost never used them in concrete situations. – Maurizio Barbato Nov 23 '22 at 13:50
  • It seems to me that an argument using the theory of field extensions splits into several cases according to whether $a/b$ is a square or even a fourth power. It will have to wait. We have many users capable of producing such an argument, so don't worry :-) – Jyrki Lahtonen Nov 23 '22 at 13:51
  • For example, in a relatively general case the polynomial $x^4-b$ is irreducible over $\Bbb{Q}$ and remains irreducible over $\Bbb{Q}(\root4\of a)$. That would give the claim because it implies that $1,\root4\of a$ and $\root4\of b$ are linearly independent over $\Bbb{Q}$ (by the proof of the tower law). But this breaks down if one of them is a square, and proving that irreducibility result may need a bit of Galois theory. More later. – Jyrki Lahtonen Nov 23 '22 at 13:55
  • @JyrkiLahtonen Dear Jyrki, the idea of using the ratio root theorem for specific values of $a$ and $b$ is a clever one! But for general values of $a$ and $b$, as I see, the argument must go through some tools of Galois theory. Moral of the story: if I want to tackle issues like this, I must familiarize much more with this wonderful mathematical theory, a true gem which the young Galois has bequeathed to all humanity! – Maurizio Barbato Nov 23 '22 at 14:06
  • @JyrkiLahtonen Dear Jyrki, just out of curiosity, what do you think is the answer in the general case of n-th roots? Is it possible that $\sqrt[n]{a}-\sqrt[n]{b}$ is rational, and so an integer, when $a$ and $b$ are two distinct positive integers such that both $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are irrational? Is this a hard problem to tackle? – Maurizio Barbato Nov 23 '22 at 15:23
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    @JyrkiLahtonen Hello. Does there exist an elementary high-school level proof for $\sqrt [4]{a}-\sqrt [4]{b}$? Is it possible? Or do we need Galois Theory? – lone student Nov 23 '22 at 17:04
  • Here is a more general version that I could easily find. I know that you are not assuming $a,b$ to be prime. That is simply a convenient case. – Jyrki Lahtonen Nov 23 '22 at 18:25
  • And, no, there are no surprise rationals of this type. A pity, in a way. I mean, it would be cool if there were, but them's the breaks. – Jyrki Lahtonen Nov 23 '22 at 18:29
  • Actually, there is the exception that with suitable choices of branches we have $$\root4\of{-4}-\sqrt{-1}=1.$$ But I'm not sure that counts here :-) – Jyrki Lahtonen Nov 23 '22 at 18:36
  • At the level of polynomials it looks like this. You see that the exception from the preceding comment is related. – Jyrki Lahtonen Nov 23 '22 at 18:39
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    @lonestudent At the moment the most interesting high school level technique I can think of would be the rational root theorem. See a few comments up. For example, by repeated squaring, we can show that $s=\root4\of7-\root4\of3$ is a zero of $$x^{16}-40x^{12}-2256x^8-27520x^4+256.$$ So by the rational root theorem we can conclude that if $s$ is rational, then it must be an integer, and also a factor of $256=2^8$. That leaves $s=1$ as the only alternative (the other factors are too large), and that is easy to rule out. – Jyrki Lahtonen Nov 23 '22 at 18:48
  • There is also this old thread. Nothing cool going on there either :-( – Jyrki Lahtonen Nov 23 '22 at 18:51
  • @JyrkiLahtonen Daer Jyrki, thank you very very ... much for your interest in my question! Your comments are really illuminating to me! – Maurizio Barbato Nov 23 '22 at 20:20

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Describing the following case as an example for I think the question deserves an answer.

Assume that $a$ has a prime factor $p$ such that it appears with an odd multiplicity in the prime factorization of $a$. It follows that $x^4-a$ is irreducible over $\Bbb{Q}$. Either by Eisenstein or its stronger cousin — the Newton's polygon. Alternatively you can look at this older question and its answers.

In particular, this means that the field extension $K/\Bbb{Q}$, $K=\Bbb{Q}(\root4\of a)$ has degree four. A basis for $K$ as a vector space over $\Bbb{Q}$ consists of the elements $\mathcal{B}=\{1,a^{1/4},a^{1/2},a^{3/4}\}$.

Next we consider the number $s=\root4\of a-\root4\of b$. We assume contrariwise that $s$ is a rational number.

Should this happen it thus follows that $\root4\of b\in K$. Furthermore, raising to the fourth power gives the equation $$ \begin{aligned} b&=(\root4\of b)^4=(\root4\of a-s)^4\\ &=a-4a^{3/4}s+6a^{1/2}s^2-4a^{1/4}s^3+s^4\\ &=(a+s^4)-4s^3a^{1/4}+6s^2a^{1/2}-4sa^{3/4}, \end{aligned} $$ where from last form we can read off the coordinates of $b$ with respect to the basis $\mathcal{B}$.

Because $b\in\Bbb{Q}$, uniqueness of the coordinates implies that all of $-4s$, $6s^2$, $-4s^3$ must vanish. In other words $s=0$, which is the uninteresting case $a=b$.

Galois theory comes to the fore, when we look at more complicated linear combinations of radicals with more than two terms, $\root4\of a+\root4\of b+\root4\of c$ and such. Mostly because it simplifies calculations of the degrees of the relevant field extensions. See the linked threads for more discussion. Even in the case of a complicated linear combination of square roots Galois theory can be used. See this old answer of mine for an example. I want to emphasize that the use of Galois theory in that answer was a bit of a cutesy show-off. We could have equally well used the natural basis over $\Bbb{Q}$ of the relevant field extension, much the same way we did here.

Jyrki Lahtonen
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  • Here I could have used the fact that as $a$ and $b$ were integers, the difference $s$ is a root of a monic polynomial $\in\Bbb{Z}[x]$, see my comments under main. It then follows that if $s$ is rational, it is actually an integer (the rational root theorem). But my argument leading to a contradiction from there would still use the properties of the basis $\mathcal{B}$, so I decided against using that. – Jyrki Lahtonen Dec 02 '22 at 11:36
  • +1. What do you think about the possible generalization $\sqrt [n]{a}-\sqrt [n]{b};?$ – lone student Dec 02 '22 at 11:51
  • @JyrkiLahtonen Dear Jyrki, I appreciate very very much that after some days in which my post attracted no interest, you have taken the time to write down the argument in full detail in a special case in order to show how we can solve the problem without using Galois theory. Magnificent! Thank you very very ... much! – Maurizio Barbato Dec 04 '22 at 18:53