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Gauss's lemma in number theory gives a condition for an integer to be a quadratic residue. Although it is not useful computationally, it has theoretical significance, being involved in some proofs of quadratic reciprocity.

It made its first appearance in Carl Friedrich Gauss's third proof (1808) of quadratic reciprocity and he proved it again in his fifth proof (1818).

Statement of the lemma

For any odd prime p let a be an integer that is coprime to p.

Consider the integers

$a,2a,3a,\dots ,\frac{p-1}{2}a$

and their least positive residues modulo p. (These residues are all distinct, so there are (p−1)/2 of them.)

Let n be the number of these residues that are greater than p/2. Then

$\left(\frac{a}{p}\right)=(-1{)}^n$

where (a/p) is the Legendre symbol.

Example

Taking p = 11 and a = 7, the relevant sequence of integers is

7, 14, 21, 28, 35.

After reduction modulo 11, this sequence becomes

7, 3, 10, 6, 2.

Three of these integers are larger than 11/2 (namely 6, 7 and 10), so n = 3. Correspondingly Gauss's lemma predicts that

$\left(\frac{7}{11}\right)=(-1{)}^3=-1.$

This is indeed correct, because 7 is not a quadratic residue modulo 11.

The above sequence of residues

7, 3, 10, 6, 2

may also be written

-4, 3, -1, -5, 2.

In this form, the integers larger than 11/2 appear as negative numbers. It is also apparent that the absolute values of the residues are a permutation of the residues

1, 2, 3, 4, 5.

Proof

A fairly simple proof of the lemma, reminiscent of one of the simplest proofs of Fermat's little theorem, can be obtained by evaluating the product

$Z=a\cdot 2a\cdot 3a\cdot \cdots \cdot \frac{p-1}{2}a$

modulo p in two different ways. On one hand it is equal to

$Z=a^{(p-1)/2}(1\cdot 2\cdot 3\cdot \cdots \cdot \frac{p-1}{2})$

The second evaluation takes more work. If x is a nonzero residue modulo p, let us define the "absolute value" of x to be

$\left|x\right|=\{\begin{array}{ll}x& \text{if }1\le x\le \frac{p-1}{2},\\ p-x& \text{if }\frac{p+1}{2}\le x\le p-1.\end{array}$

Since n counts those multiples ka which are in the latter range, and since for those multiples, −ka is in the first range, we have

$Z=(-1{)}^n\left(\right|a|\cdot |2a|\cdot |3a|\cdot \cdots \cdots \left|\frac{p-1}{2}a\right|).$

Now observe that the values |ra| are distinct for r = 1, 2, ..., (p−1)/2. Indeed, if |ra| = |sa|, then ra = ±sa, and therefore r = ±s (because a is invertible modulo p), so r = s because they are both in the range 1 ≤ r ≤ (p−1)/2. But there are exactly (p−1)/2 of them, so they must just be some rearrangement of the integers 1, 2, ..., (p−1)/2. Therefore

$Z=(-1{)}^n(1\cdot 2\cdot 3\cdot \cdots \cdot \frac{p-1}{2}).$

Comparing with our first evaluation, we may cancel out the nonzero factor

$1\cdot 2\cdot 3\cdot \cdots \cdot \frac{p-1}{2}$

and we are left with

$a^{(p-1)/2}=(-1{)}^n.$

This is the desired result, because by Euler's criterion the left hand side is just an alternative expression for the Legendre symbol (a/p).

Applications

Gauss's lemma is used in many, but by no means all, of the known proofs of quadratic reciprocity.

For example, Eisenstein used Gauss's lemma to prove that if p is an odd prime then

$\left(\frac{a}{p}\right)=\prod _{n=1}^{(p-1)/2}\frac{\sin (2\pi an/p)}{\sin (2\pi n/p)},$

and used this formula to prove quadratic reciprocity, (and, by using elliptic rather than circular functions, to prove the cubic and quartic reciprocity laws.)

Kronecker used the lemma to show that

$\left(\frac{p}{q}\right)=\mathrm{sgn}\prod _{i=1}^{\frac{q-1}{2}}\prod _{k=1}^{\frac{p-1}{2}}(\frac{k}{p}-\frac{i}{q}).$

Switching p and q immediately gives quadratic reciprocity.

It is also used in what are probably the simplest proofs of the "second supplementary law"

$\left(\frac{2}{p}\right)=(-1{)}^{(p^2-1)/8}=\{\begin{array}{l}+1\text{ if }p\equiv \pm 1(\mathrm{mod}8)\\ -1\text{ if }p\equiv \pm 3(\mathrm{mod}8)\end{array}$

Higher powers

Generalizations of Gauss's lemma can be used to compute higher power residue symbols. In his second monograph on biquadratic reciprocity, Gauss used a fourth-power lemma to derive the formula for the biquadratic character of 1 + i in Z[i], the ring of Gaussian integers. Subsequently, Eisenstein used third- and fourth-power versions to prove cubic and quartic reciprocity.

nth power residue symbol

Mining Engineer (Excluding Oil ) Truman from Alma, loves to spend time knotting, largest property developers in singapore developers in singapore and stamp collecting. Recently had a family visit to Urnes Stave Church. Let k be an algebraic number field with ring of integers   ${\mathcal{𝒪}}_k,$   and let   $\mathfrak{p}\subset {\mathcal{𝒪}}_k$   be a prime ideal. The ideal norm of  $\mathfrak{p}$  is defined as the cardinality of the residue class ring (since $\mathfrak{p}$ is prime this is a finite field)   ${\mathcal{𝒪}}_k/\mathfrak{p}:\mathrm{N}\mathfrak{p}=|{\mathcal{𝒪}}_k/\mathfrak{p}|.$

Assume that a primitive nth root of unity   ${\zeta }_n\in {\mathcal{𝒪}}_k,$   and that n and $\mathfrak{p}$ are coprime (i.e. $n\in ̸\mathfrak{p}.$)   Then

No two distinct nth roots of unity can be congruent $(\mathrm{mod}\mathfrak{p}).$

The proof is by contradiction: assume otherwise, that   ${\zeta }_n^r\equiv {\zeta }_n^s(\mathrm{mod}\mathfrak{p}),0<r<s\le n.$   Then letting   $t=s-r,{\zeta }_n^t\equiv 1(\mathrm{mod}\mathfrak{p}),$   and  $0<t<n.$ From the definition of roots of unity,

$x^n-1=(x-1)(x-{\zeta }_n)(x-{\zeta }_n^2)\dots (x-{\zeta }_n^{n-1}),$   and dividing by x − 1 gives

$x^{n-1}+x^{n-2}+\dots +x+1=(x-{\zeta }_n)(x-{\zeta }_n^2)\dots (x-{\zeta }_n^{n-1}).$

Letting x = 1 and taking residues $(\mathrm{mod}\mathfrak{p}),$

$n\equiv (1-{\zeta }_n)(1-{\zeta }_n^2)\dots (1-{\zeta }_n^{n-1})(\mathrm{mod}\mathfrak{p}).$

Since n and $\mathfrak{p}$  are coprime, $n\equiv ̸0(\mathrm{mod}\mathfrak{p}),$   but under the assumption, one of the factors on the right must be zero. Therefore the assumption that two distinct roots are congruent is false.

Thus the residue classes of   ${\mathcal{𝒪}}_k/\mathfrak{p}$   containing the powers of ζ_n are a subgroup of order n of its (multiplicative) group of units,   $({\mathcal{𝒪}}_k/\mathfrak{p}{)}^{\times }={\mathcal{𝒪}}_k/\mathfrak{p}-\{0\}.$   Therefore the order of   $({\mathcal{𝒪}}_k/\mathfrak{p}{)}^{\times }$   is a multiple of n, and

$\mathrm{N}\mathfrak{p}=|{\mathcal{𝒪}}_k/\mathfrak{p}|=|({\mathcal{𝒪}}_k/\mathfrak{p}{)}^{\times }|+1\equiv 1(\mathrm{mod}n).$

There is an analogue of Fermat's theorem in   ${\mathcal{𝒪}}_k:$  If   $\alpha \in {\mathcal{𝒪}}_k,\alpha \in ̸\mathfrak{p},$   then

${\alpha }^{\mathrm{N}\mathfrak{p}-1}\equiv 1(\mathrm{mod}\mathfrak{p}),$   and since   $\mathrm{N}\mathfrak{p}\equiv 1(\mathrm{mod}n),$

${\alpha }^{\frac{\mathrm{N}\mathfrak{p}-1}{n}}\equiv {\zeta }_n^s(\mathrm{mod}\mathfrak{p})$   is well-defined and congruent to a unique nth root of unity ζ_n^s.

This root of unity is called the nth-power residue symbol for   ${\mathcal{𝒪}}_k,$   and is denoted by

${\left(\frac{\alpha }{\mathfrak{p}}\right)}_n={\zeta }_n^s\equiv {\alpha }^{\frac{\mathrm{N}\mathfrak{p}-1}{n}}(\mathrm{mod}\mathfrak{p}).$

It can be proven that

${\left(\frac{\alpha }{\mathfrak{p}}\right)}_n=1\text{ if and only if there is an }\eta \in {\mathcal{𝒪}}_k\text{ such that }\alpha \equiv {\eta }^n(\mathrm{mod}\mathfrak{p}).$

1/n systems

Let   ${\mu }_n=\{1,{\zeta }_n,{\zeta }_n^2,\dots ,{\zeta }_n^{n-1}\}$   be the multiplicative group of the nth roots of unity, and let   $A=\{a_1,a_2,\dots ,a_m\}$   be representatives of the cosets of   $({\mathcal{𝒪}}_k/\mathfrak{p}{)}^{\times }/{\mu }_n.$   Then A is called a 1/n system $(\mathrm{mod}\mathfrak{p}).$

In other words, there are  $mn=\mathrm{N}\mathfrak{p}-1$  numbers in the set   $A\mu =\{a_i{\zeta }_n^j:1\le i\le m,0\le j\le n-1\},$   and this set constitutes a representative set for   $({\mathcal{𝒪}}_k/\mathfrak{p}{)}^{\times }.$

The numbers 1, 2, ..., (p − 1)/2, used in the original version of the lemma, are a 1/2 system (mod p).

Constructing a 1/n system is straightforward: let M be a representative set for   $({\mathcal{𝒪}}_k/\mathfrak{p}{)}^{\times }.$   Pick any $a_1\in M$   and remove the numbers congruent to  $a_1,a_1{\zeta }_n,a_1{\zeta }_n^2,\dots ,a_1{\zeta }_n^{n-1}$   from M. Pick a₂ from M and remove the numbers congruent to   $a_2,a_2{\zeta }_n,a_2{\zeta }_n^2,\dots ,a_2{\zeta }_n^{n-1}$   Repeat until M is exhausted. Then {a₁, a₂, ... a_m} is a 1/n system $(\mathrm{mod}\mathfrak{p}).$

The lemma for nth powers

Gauss's lemma for the nth power residue symbol is

Let   ${\zeta }_n\in {\mathcal{𝒪}}_k$   be a primitive nth root of unity,   $\mathfrak{p}\subset {\mathcal{𝒪}}_k$   a prime ideal,   $\gamma \in {\mathcal{𝒪}}_k,n\gamma \in ̸\mathfrak{p},$ (i.e. $\mathfrak{p}$ is coprime to both γ and n) and let A = {a₁, a₂,..., a_m} be a 1/n system $(\mathrm{mod}\mathfrak{p}).$

Then for each i, 1 ≤ i ≤ m, there are integers π(i), unique (mod m), and b(i), unique (mod n), such that

$\gamma a_i\equiv {\zeta }_n^{b(i)}{a}_{\pi (i)}(\mathrm{mod}\mathfrak{p}),$

and the nth-power residue symbol is given by the formula

${\left(\frac{\gamma }{\mathfrak{p}}\right)}_n={\zeta }_n^{b(1)+b(2)+\dots +b(m)}.$

The classical lemma for the quadratic Legendre symbol is the special case n = 2, ζ₂ = −1, A = {1, 2, ..., (p − 1)/2}, b(k) = 1 if ak > p/2, b(k) = 0 if ak < p/2.

Proof

The proof of the nth-power lemma uses the same ideas that were used in the proof of the quadratic lemma.

The existence of the integers π(i) and b(i), and their uniqueness (mod m) and (mod n), respectively, come from the fact that Aμ is a representative set.

Assume that π(i) = π(j) = p, i.e.

$\gamma a_i\equiv {\zeta }_n^r{a}_p(\mathrm{mod}\mathfrak{p})$   and   $\gamma a_j\equiv {\zeta }_n^s{a}_p(\mathrm{mod}\mathfrak{p}).$

Then

${\zeta }_n^{s-r}\gamma a_i\equiv {\zeta }_n^s{a}_p\equiv \gamma a_j(\mathrm{mod}\mathfrak{p})$

Because γ and $\mathfrak{p}$ are coprime both sides can be divided by γ, giving

${\zeta }_n^{s-r}{a}_i\equiv a_j(\mathrm{mod}\mathfrak{p}),$

which, since A is a 1/n system, implies s = r and i = j, showing that π is a permutation of the set {1, 2, ..., m}.

Then on the one hand, by the definition of the power residue symbol,

$\begin{array}{rl}(\gamma a_1)(\gamma a_2)\dots (\gamma a_m)& ={\gamma }^{\frac{\mathrm{N}\mathfrak{p}-1}{n}}{a}_1{a}_2\dots a_m\\ & \equiv {\left(\frac{\gamma }{\mathfrak{p}}\right)}_n{a}_1{a}_2\dots a_m(\mathrm{mod}\mathfrak{p}),\end{array}$

and on the other hand, since π is a permutation,

$\begin{array}{rl}(\gamma a_1)(\gamma a_2)\dots (\gamma a_m)& \equiv {\zeta }_n^{b(1)}{a}_{\pi (1)}{\zeta }_n^{b(2)}{a}_{\pi (2)}\dots {\zeta }_n^{b(m)}{a}_{\pi (m)}\\ & \equiv {\zeta }_n^{b(1)+b(2)+\dots +b(m)}{a}_{\pi (1)}{a}_{\pi (2)}\dots a_{\pi (m)}\\ & \equiv {\zeta }_n^{b(1)+b(2)+\dots +b(m)}{a}_1{a}_2\dots a_m(\mathrm{mod}\mathfrak{p}),\end{array}$

so

${\left(\frac{\gamma }{\mathfrak{p}}\right)}_n{a}_1{a}_2\dots a_m\equiv {\zeta }_n^{b(1)+b(2)+\dots +b(m)}{a}_1{a}_2\dots a_m(\mathrm{mod}\mathfrak{p}),$

and since for all 1 ≤ i ≤ m, a_i and  $\mathfrak{p}$   are coprime, a₁a₂...a_m can be cancelled from both sides of the congruence,

${\left(\frac{\gamma }{\mathfrak{p}}\right)}_n\equiv {\zeta }_n^{b(1)+b(2)+\dots +b(m)}(\mathrm{mod}\mathfrak{p}),$

and the theorem follows from the fact that no two distinct n^th roots of unity can be congruent (mod $\mathfrak{p}$).

Relation to the transfer in group theory

Let G be the multiplicative group of nonzero residue classes in Z/pZ, and let H be the subgroup {+1, −1}. Consider the following coset representatives of H in G,

$1,2,3,\dots ,\frac{p-1}{2}.$

Applying the machinery of the transfer to this collection of coset representatives, we obtain the transfer homomorphism

$\varphi :G\to H,$

which turns out to be the map that sends a to (−1)ⁿ, where a and n are as in the statement of the lemma. Gauss's lemma may then be viewed as a computation that explicitly identifies this homomorphism as being the quadratic residue character.

See also

Two other characterizations of squares modulo a prime are Euler's criterion and Zolotarev's lemma.

Notes

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References

The two monographs Gauss published on biquadratic reciprocity have consecutively numbered sections: the first contains §§ 1–23 and the second §§ 24–76. Footnotes referencing these are of the form "Gauss, BQ, § n".

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  Naturally, you will have to pay a safety deposit and that is usually one month rent for annually of the settlement. That is the place your good religion deposit will likely be taken into account and will kind part or all of your security deposit. Anticipate to have a proportionate amount deducted out of your deposit if something is discovered to be damaged if you move out. It's best to you'll want to test the inventory drawn up by the owner, which can detail all objects in the property and their condition. If you happen to fail to notice any harm not already mentioned within the inventory before transferring in, you danger having to pay for it yourself.
  
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  Have handed an trade examination i.e Widespread Examination for House Brokers (CEHA) or Actual Property Agency (REA) examination, or equal; Exclusive brokers are extra keen to share listing information thus making certain the widest doable coverage inside the real estate community via Multiple Listings and Networking. Accepting a severe provide is simpler since your agent is totally conscious of all advertising activity related with your property. This reduces your having to check with a number of agents for some other offers. Price control is easily achieved. Paint work in good restore-discuss with your Property Marketing consultant if main works are still to be done. Softening in residential property prices proceed, led by 2.8 per cent decline within the index for Remainder of Central Region
  
  Once you place down the one per cent choice price to carry down a non-public property, it's important to accept its situation as it is whenever you move in – faulty air-con, choked rest room and all. Get round this by asking your agent to incorporate a ultimate inspection clause within the possibility-to-buy letter. HDB flat patrons routinely take pleasure in this security net. "There's a ultimate inspection of the property two days before the completion of all HDB transactions. If the air-con is defective, you can request the seller to repair it," says Kelvin.
  
  15.6.1 As the agent is an intermediary, generally, as soon as the principal and third party are introduced right into a contractual relationship, the agent drops out of the image, subject to any problems with remuneration or indemnification that he could have against the principal, and extra exceptionally, against the third occasion. Generally, agents are entitled to be indemnified for all liabilities reasonably incurred within the execution of the brokers´ authority.
  
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These are in Gauss's Werke, Vol II, pp. 65–92 and 93–148

German translations of the above are in the following, which also has the Disquisitiones Arithmeticae and Gauss's other papers on number theory, including the six proofs of quadratic reciprocity.

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  Naturally, you will have to pay a safety deposit and that is usually one month rent for annually of the settlement. That is the place your good religion deposit will likely be taken into account and will kind part or all of your security deposit. Anticipate to have a proportionate amount deducted out of your deposit if something is discovered to be damaged if you move out. It's best to you'll want to test the inventory drawn up by the owner, which can detail all objects in the property and their condition. If you happen to fail to notice any harm not already mentioned within the inventory before transferring in, you danger having to pay for it yourself.
  
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  Have handed an trade examination i.e Widespread Examination for House Brokers (CEHA) or Actual Property Agency (REA) examination, or equal; Exclusive brokers are extra keen to share listing information thus making certain the widest doable coverage inside the real estate community via Multiple Listings and Networking. Accepting a severe provide is simpler since your agent is totally conscious of all advertising activity related with your property. This reduces your having to check with a number of agents for some other offers. Price control is easily achieved. Paint work in good restore-discuss with your Property Marketing consultant if main works are still to be done. Softening in residential property prices proceed, led by 2.8 per cent decline within the index for Remainder of Central Region
  
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  15.6.1 As the agent is an intermediary, generally, as soon as the principal and third party are introduced right into a contractual relationship, the agent drops out of the image, subject to any problems with remuneration or indemnification that he could have against the principal, and extra exceptionally, against the third occasion. Generally, agents are entitled to be indemnified for all liabilities reasonably incurred within the execution of the brokers´ authority.
  
  To achieve the very best outcomes, you must be always updated on market situations, including past transaction information and reliable projections. You could review and examine comparable homes that are currently available in the market, especially these which have been sold or not bought up to now six months. You'll be able to see a pattern of such report by clicking here It's essential to defend yourself in opposition to unscrupulous patrons. They are often very skilled in using highly unethical and manipulative techniques to try and lure you into a lure. That you must also protect your self, your loved ones, and personal belongings as you'll be serving many strangers in your home. Sign a listing itemizing of all of the objects provided by the proprietor, together with their situation. HSR Prime Recruiter 2010
• Many property agents need to declare for the PIC grant in Singapore. However, not all of them know find out how to do the correct process for getting this PIC scheme from the IRAS. There are a number of steps that you need to do before your software can be approved.
  
  Naturally, you will have to pay a safety deposit and that is usually one month rent for annually of the settlement. That is the place your good religion deposit will likely be taken into account and will kind part or all of your security deposit. Anticipate to have a proportionate amount deducted out of your deposit if something is discovered to be damaged if you move out. It's best to you'll want to test the inventory drawn up by the owner, which can detail all objects in the property and their condition. If you happen to fail to notice any harm not already mentioned within the inventory before transferring in, you danger having to pay for it yourself.
  
  In case you are in search of an actual estate or Singapore property agent on-line, you simply should belief your intuition. It's because you do not know which agent is nice and which agent will not be. Carry out research on several brokers by looking out the internet. As soon as if you end up positive that a selected agent is dependable and reliable, you can choose to utilize his partnerise in finding you a home in Singapore. Most of the time, a property agent is taken into account to be good if he or she locations the contact data on his website. This may mean that the agent does not mind you calling them and asking them any questions relating to new properties in singapore in Singapore. After chatting with them you too can see them in their office after taking an appointment.
  
  Have handed an trade examination i.e Widespread Examination for House Brokers (CEHA) or Actual Property Agency (REA) examination, or equal; Exclusive brokers are extra keen to share listing information thus making certain the widest doable coverage inside the real estate community via Multiple Listings and Networking. Accepting a severe provide is simpler since your agent is totally conscious of all advertising activity related with your property. This reduces your having to check with a number of agents for some other offers. Price control is easily achieved. Paint work in good restore-discuss with your Property Marketing consultant if main works are still to be done. Softening in residential property prices proceed, led by 2.8 per cent decline within the index for Remainder of Central Region
  
  Once you place down the one per cent choice price to carry down a non-public property, it's important to accept its situation as it is whenever you move in – faulty air-con, choked rest room and all. Get round this by asking your agent to incorporate a ultimate inspection clause within the possibility-to-buy letter. HDB flat patrons routinely take pleasure in this security net. "There's a ultimate inspection of the property two days before the completion of all HDB transactions. If the air-con is defective, you can request the seller to repair it," says Kelvin.
  
  15.6.1 As the agent is an intermediary, generally, as soon as the principal and third party are introduced right into a contractual relationship, the agent drops out of the image, subject to any problems with remuneration or indemnification that he could have against the principal, and extra exceptionally, against the third occasion. Generally, agents are entitled to be indemnified for all liabilities reasonably incurred within the execution of the brokers´ authority.
  
  To achieve the very best outcomes, you must be always updated on market situations, including past transaction information and reliable projections. You could review and examine comparable homes that are currently available in the market, especially these which have been sold or not bought up to now six months. You'll be able to see a pattern of such report by clicking here It's essential to defend yourself in opposition to unscrupulous patrons. They are often very skilled in using highly unethical and manipulative techniques to try and lure you into a lure. That you must also protect your self, your loved ones, and personal belongings as you'll be serving many strangers in your home. Sign a listing itemizing of all of the objects provided by the proprietor, together with their situation. HSR Prime Recruiter 2010