# 'Euler criterion' for cube roots?

• Camila86
In summary, the conversation discusses deriving Euler's criterion for the existence of cube roots modulo p, prime. The cases for p=3k+2 and p=3k+1 are discussed, with the latter being more complex due to 3 dividing p-1. It is shown that for p=3k+2, every element has a cube root, while for p=3k+1, there are (p-1)/3 solutions that can be shown to be cubes. It is also mentioned that each cube has three roots, and a solution for this is given using the cube root of unity.

#### Camila86

I am trying to derive a version of Euler's criterion for the existence of cube roots modulo p, prime.

So far, I have split the primes up into two cases:

For p = 3k+2, every a(mod p) has a cube root.

For p = 3k+1, I don't know which a it is true for, but I did a few examples and noticed a couple of things:

* If a is a cube, so is -a
* If a is a cube, it has exactly 3 cube roots

Does this have anything to do with p-1= 0(mod3)?

For the form p=3k+2, three does not divide p-1, the order of the group, and so cubing the elements just permutates the residues, so every element (trivially) has a cube root.

Now in the case where 3 divides p-1, we have X^(p-1)/3 -1==0 Mod p and it will have (p-1)/3 elements that satisfy it. Proof:

X^(p-1)-1 =={X^(p-1)/3-1}{X^2(p-1)/3+X^(p-1)/3+1} Mod p, and since the left side has p-1 distinct solutions, (the entire multiplicative residue set), so does the other side and thus there are (p-1)/3 solutions to the first part of the right side. All of these can be shown to be cubes. Proof:

Let b^(p-1)/3 ==1 Mod p. b, itself, is a power of a primative root b=d^u. Since d^u(p-1)/3==1 Mod p, and p-1 is the smallest positive power such that this occurs it follows that u/3 is an integer, and so b is a cube.

Each cube has three roots. Proof: Firstly with all elements cubed we know that 2/3 of these cubes are duplicates. Give a^3==b^3 Mod p, either a==b or a^2+ab+b^2==0 Mod p. Here we make a transformation used by Euler: Let x+y=2a, x-y=2b, then the above equation transforms (1/4)(3X^2+Y^2==0 Mod p), thus there is a solution to (Y/X)^2== -3 Mod p. So there is a solution to

$$\beta=\frac{-1+\sqrt-3}{2}$$ This element represents the cube root of unity and so given a^3, have the three roots: $$a,a\beta, a\beta^2$$

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The Euler criterion is a theorem in number theory that provides a necessary and sufficient condition for a quadratic residue to exist modulo a prime number. It states that if a is a quadratic residue modulo p, where p is an odd prime, then a^((p-1)/2) ≡ 1 (mod p). This criterion has been extended to higher powers, such as the Legendre symbol for cubic residues.

In your case, you are trying to derive a version of Euler's criterion for the existence of cube roots modulo p, where p is an odd prime. It is important to note that this version of the criterion is not a generalization of the original criterion, but rather a separate theorem.

Your approach of splitting the primes into two cases is a good start. For p = 3k+2, it is true that every a (mod p) has a cube root. This can be easily proven by using the fact that (p-1)/3 is an integer for p = 3k+2, and thus a^((p-1)/3) ≡ 1 (mod p) for any a (mod p).

For p = 3k+1, where p is congruent to 1 modulo 3, things get a bit more complicated. As you have noticed, if a is a cube, then so is -a. This is because (-a)^3 ≡ -(a^3) ≡ -a (mod p). However, this does not mean that every a (mod p) has a cube root.

In fact, it can be proven that if a is a cube (mod p), then it has exactly 3 cube roots. This can be seen by considering the numbers a^((p+1)/3), a^((p+1)/3) * ω, and a^((p+1)/3) * ω^2, where ω is a primitive cube root of unity modulo p. These three numbers are distinct and are all cubes (mod p), thus giving us the three cube roots of a (mod p).

To answer your question about the relationship between p-1 ≡ 0 (mod 3) and the existence of cube roots, it is worth noting that p-1 ≡ 0 (mod 3) implies that p is congruent to 1 modulo 3. This means that for any a (mod p), there will be

## 1. What is the Euler criterion for cube roots?

The Euler criterion for cube roots is a mathematical rule used to determine if a number is a perfect cube. It states that if the sum of the digits of a number is divisible by 3, then the number is a perfect cube.

## 2. How is the Euler criterion for cube roots used?

The Euler criterion for cube roots is used by first calculating the sum of the digits of a given number. If the sum is divisible by 3, then the number is a perfect cube and its cube root can be easily calculated. If the sum is not divisible by 3, then the number is not a perfect cube.

## 3. What is an example of using the Euler criterion for cube roots?

For example, let's say we have the number 216. The sum of its digits is 2+1+6 = 9, which is divisible by 3. Therefore, 216 is a perfect cube and its cube root is 6. On the other hand, if we have the number 245, the sum of its digits is 2+4+5 = 11, which is not divisible by 3. Therefore, 245 is not a perfect cube.

## 4. Is the Euler criterion for cube roots always accurate?

Yes, the Euler criterion for cube roots is always accurate. However, it only works for perfect cubes and may not work for numbers that are not perfect cubes. Therefore, it is important to double check the result.

## 5. What is the significance of the Euler criterion for cube roots?

The Euler criterion for cube roots is significant because it provides a quick and easy way to determine if a number is a perfect cube without having to actually calculate its cube root. This can be useful in many mathematical and scientific calculations that involve perfect cubes.