Euler's criterion

In number theory, Euler's criterion is a formula for determining whether an integer is a quadratic residue modulo a prime. Precisely,

Let p be an odd prime and a be an integer coprime to p. Then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{\tfrac{p-1}{2}} \equiv \begin{cases} \;\;\,1\pmod{p}& \text{ if there is an integer }x \text{ such that }x^2\equiv a \pmod{p},\\ -1\pmod{p}& \text{ if there is no such integer.} \end{cases} }

Euler's criterion can be concisely reformulated using the Legendre symbol:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{a}{p}\right) \equiv a^{\tfrac{p-1}{2}} \pmod p. }

The criterion dates from a 1748 paper by Leonhard Euler.

Proof

The proof uses the fact that the residue classes modulo a prime number are a field. See the article prime field for more details.

Because the modulus is prime, Lagrange's theorem applies: a polynomial of degree k can only have at most k roots. In particular, x² ≡ a (mod p) has at most 2 solutions for each a. This immediately implies that besides 0 there are at least p − 1/2 distinct quadratic residues modulo p: each of the p − 1 possible values of x can only be accompanied by one other to give the same residue.

In fact, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (p-x)^{2}\equiv x^{2} \pmod p.} This is because Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (p-x)^{2} \equiv p^{2}-{2}{x}{p}+x^{2} \equiv x^{2} \pmod p.} So, the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{p-1}{2}} distinct quadratic residues are: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1^{2}, 2^{2}, ... , (\tfrac{p-1}{2})^{2} \pmod p. }

As a is coprime to p, Fermat's little theorem says that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{p-1}\equiv 1 \pmod p, }

which can be written as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( a^{\tfrac{p-1}{2}}-1 \right)\left( a^{\tfrac{p-1}{2}}+1 \right) \equiv 0 \pmod p. }

Since the integers mod p form a field, for each a, one or the other of these factors must be zero. Therefore,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{\tfrac{p-1}{2}}\equiv 1\pmod p } or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{\tfrac{p-1}{2}} \equiv {-1}\pmod p. }

Now if a is a quadratic residue, a ≡ x²,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{\tfrac{p-1}{2}}\equiv {(x^2)}^{\tfrac{p-1}{2}} \equiv x^{p-1}\equiv1\pmod p. }

So every quadratic residue (mod p) makes the first factor zero.

Applying Lagrange's theorem again, we note that there can be no more than p − 1/2 values of a that make the first factor zero. But as we noted at the beginning, there are at least p − 1/2 distinct quadratic residues (mod p) (besides 0). Therefore, they are precisely the residue classes that make the first factor zero. The other p − 1/2 residue classes, the nonresidues, must make the second factor zero, or they would not satisfy Fermat's little theorem. This is Euler's criterion.

Alternative proof

This proof only uses the fact that any congruence Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle kx\equiv l\!\!\! \pmod p} has a unique (modulo Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} ) solution Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} provided Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} does not divide Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} . (This is true because as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} runs through all nonzero remainders modulo Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} without repetitions, so does Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle kx} : if we have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle kx_1\equiv kx_2 \pmod p} , then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p\mid k(x_1-x_2)} , hence $p\mid (x_{1}-x_{2})$, but Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_1} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_2} aren't congruent modulo Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} .) It follows from this fact that all nonzero remainders modulo Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} the square of which isn't congruent to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} can be grouped into unordered pairs Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x,y)} according to the rule that the product of the members of each pair is congruent to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} modulo Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} (since by this fact for every Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} we can find such an Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} , uniquely, and vice versa, and they will differ from each other if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y^2} is not congruent to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} ). If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} is not a quadratic residue, this is simply a regrouping of all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p-1} nonzero residues into Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (p-1)/2} pairs, hence we conclude that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1\cdot2\cdot ... \cdot (p-1)\equiv a^{\frac{p-1}{2}} \!\!\! \pmod p} . If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} is a quadratic residue, exactly two remainders were not among those paired, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -r} such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r^2\equiv a\!\!\! \pmod p} . If we pair those two absent remainders together, their product will be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -a} rather than Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} , whence in this case Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1\cdot2\cdot ... \cdot (p-1)\equiv -a^{\frac{p-1}{2}} \!\!\! \pmod p} . In summary, considering these two cases we have demonstrated that for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a\not\equiv 0 \!\!\! \pmod p} we have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1\cdot2\cdot ... \cdot (p-1)\equiv -\left(\frac{a}{p}\right)a^{\frac{p-1}{2}} \!\!\! \pmod p} . It remains to substitute Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a=1} (which is obviously a square) into this formula to obtain at once Wilson's theorem, Euler's criterion, and (by squaring both sides of Euler's criterion) Fermat's little theorem.

Examples

Example 1: Finding primes for which a is a residue

Let a = 17. For which primes p is 17 a quadratic residue?

We can test prime p's manually given the formula above.

In one case, testing p = 3, we have 17^{(3 − 1)/2} = 17¹ ≡ 2 ≡ −1 (mod 3), therefore 17 is not a quadratic residue modulo 3.

In another case, testing p = 13, we have 17^{(13 − 1)/2} = 17⁶ ≡ 1 (mod 13), therefore 17 is a quadratic residue modulo 13. As confirmation, note that 17 ≡ 4 (mod 13), and 2² = 4.

We can do these calculations faster by using various modular arithmetic and Legendre symbol properties.

If we keep calculating the values, we find:

(17/p) = +1 for p = {13, 19, ...} (17 is a quadratic residue modulo these values)

(17/p) = −1 for p = {3, 5, 7, 11, 23, ...} (17 is not a quadratic residue modulo these values).

Example 2: Finding residues given a prime modulus p

Which numbers are squares modulo 17 (quadratic residues modulo 17)?

We can manually calculate it as:

1² = 1

2² = 4

3² = 9

4² = 16

5² = 25 ≡ 8 (mod 17)

6² = 36 ≡ 2 (mod 17)

7² = 49 ≡ 15 (mod 17)

8² = 64 ≡ 13 (mod 17).

So the set of the quadratic residues modulo 17 is {1,2,4,8,9,13,15,16}. Note that we did not need to calculate squares for the values 9 through 16, as they are all negatives of the previously squared values (e.g. 9 ≡ −8 (mod 17), so 9² ≡ (−8)² = 64 ≡ 13 (mod 17)).

We can find quadratic residues or verify them using the above formula. To test if 2 is a quadratic residue modulo 17, we calculate 2^{(17 − 1)/2} = 2⁸ ≡ 1 (mod 17), so it is a quadratic residue. To test if 3 is a quadratic residue modulo 17, we calculate 3^{(17 − 1)/2} = 3⁸ ≡ 16 ≡ −1 (mod 17), so it is not a quadratic residue.

Euler's criterion is related to the law of quadratic reciprocity.

Applications

In practice, it is more efficient to use an extended variant of Euclid's algorithm to calculate the Jacobi symbol Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{a}{n}\right)} . If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} is an odd prime, this is equal to the Legendre symbol, and decides whether Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} is a quadratic residue modulo Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} .

On the other hand, since the equivalence of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^\frac{n-1}{2}} to the Jacobi symbol holds for all odd primes, but not necessarily for composite numbers, calculating both and comparing them can be used as a primality test, specifically the Solovay–Strassen primality test. Composite numbers for which the congruence holds for a given Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} are called Euler–Jacobi pseudoprimes to base Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} .

References

The Disquisitiones Arithmeticae has been translated from Gauss's Ciceronian Latin into English and German. The German edition includes all of his papers on number theory: all the proofs of quadratic reciprocity, the determination of the sign of the Gauss sum, the investigations into biquadratic reciprocity, and unpublished notes.

• Gauss, Carl Friedrich (1986), Disquisitiones Arithemeticae (Second, corrected edition), translated by Clarke, Arthur A. (English), New York: Springer, ISBN 0-387-96254-9

• Gauss, Carl Friedrich (1965), Untersuchungen über höhere Arithmetik (Disquisitiones Arithemeticae & other papers on number theory) (Second edition), translated by Maser, H. (German), New York: Chelsea, ISBN 0-8284-0191-8

• Hardy, G. H.; Wright, E. M. (1980), An Introduction to the Theory of Numbers (Fifth edition), Oxford: Oxford University Press, ISBN 978-0-19-853171-5

• Lemmermeyer, Franz (2000), Reciprocity Laws: from Euler to Eisenstein, Berlin: Springer, ISBN 3-540-66957-4

External links

• The Euler Archive