[Math] What rational numbers have rational square roots

elementary-number-theoryradicalsrationality-testing

All rational numbers have the fraction form $$\frac a b,$$ where a and b are integers($b\neq0$).

My question is: for what $a$ and $b$ does the fraction have rational square root? The simple answer would be when both are perfect squares, but if two perfect squares are multiplied by a common integer $n$, the result may not be two perfect squares. Like:$$\frac49 \to \frac 8 {18}$$

And intuitively, without factoring, $a=8$ and $b=18$ must qualify by some standard to have a rational square root.

Once this is solved, can this be extended to any degree of roots? Like for what $a$ and $b$ does the fraction have rational $n$th root?

Best Answer

A nice generalization of the fundamental theorem of arithmetic is that every rational number is uniquely represented as a product of primes raised to integer powers. For example:

$$\frac{4}{9} = 2^{2}*3^{-2}$$

This is the natural generalization of factoring integers to rational numbers. Positive powers are part of the numerator, negative powers part of the denominator (since $a^{-b} = \frac{1}{a^b}$).

When you take the $n$th root, you divide each power by $n$:

$$\sqrt[n]{2^{p_2}*3^{p_3}*5^{p_5}...} = 2^{p_2/n}*3^{p_3/n}*5^{p_5/n}...$$

For example:

$$\sqrt{\frac{4}{9}} = 2^{2/2}*3^{-2/2} = \frac{2}{3}$$

In order for the powers to continue being integers when we divide (and thus the result a rational number), they must be multiples of $n$. In the case where $n$ is $2$, that means the numerator and denominator, in their reduced form, are squares. (And for $n=3$, cubes, and so on...)

In your example, when you multiply the numerator and denominator by the same number, they continue to be the same rational number, just represented differently.

$$\frac{2*4}{2*9} = 2^{2+1-1}*3^{-2} = 2^{2}*3^{-2}$$

You correctly recognize the important of factoring, though you don't really want to use it in your answer. But the most natural way to test if the fraction produced by dividing $a$ by $b$ has a rational $n$th root, is to factor $a/b$ and look at the powers. Or, equivalently, reduce the fraction and determine if the numerator and denominator are integers raised to the power of $n$.

Related Solutions

[Math] Writing square root of square-free numbers as sum of square roots.

Lemma 1. If $m$ is a positive integer and $\sqrt m$ is rational, then $\sqrt m$ is an integer.

Proof. Easy.

Lemma 2. If $m,n$ are positive integers and $\sqrt m+\sqrt n$ is rational, then both $\sqrt m$ and $\sqrt n$ are integers.

Proof. Say $\sqrt m+\sqrt n=x\in\Bbb Q$. Then $$\sqrt m-\sqrt n=\frac{m-n}{x}$$ is rational and so is $$\sqrt m=\frac{(\sqrt m+\sqrt n)+(\sqrt m-\sqrt n)}{2}\ ,$$ and likewise $\sqrt n\,$. By lemma 1, $\sqrt m$ and $\sqrt n$ are integers.

Now suppose that $$\sqrt a+\sqrt b+\sqrt c=\sqrt s\ ,$$ where $a,b,c,s$ are positive integers and $s$ is squarefree. Squaring and rearranging, $$2\bigl(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\bigl)=s-a-b-c\ .$$ Now add to this equation the identity $2\sqrt a\sqrt a=2a$ and factorise to obtain $$2\sqrt{bc}+2\sqrt{as}=s+a-b-c\ .$$ By lemma 2, we see that $\sqrt{as}$ is an integer; since $s$ is squarefree, $a$ must be a square times $s$, say $a=p^2s$. Similarly $b=q^2s$ and $c=r^2s$, so $$p\sqrt s+q\sqrt s+r\sqrt s=\sqrt s\ ,$$ but as $p+q+r>1$, this is impossible.

[Math] Perfect square palindromic numbers

This is not an answer!

The examples below a million are :

26 676
264 69696
307 94249
836 698896
2285 5221225
2636 6948496
22865 522808225
24846 617323716
30693 942060249
798644 637832238736

Best Answer

Related Solutions

[Math] Writing square root of square-free numbers as sum of square roots.

[Math] Perfect square palindromic numbers

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