5.5: Legendre Symbol
In this section, we define Legendre symbol which is a notation associated to quadratic residues and prove related theorems.
Definition: Legendre symbol
Let \(p\neq 2\) be a prime and \(a\) be an integer such that \(p\nmid a\). The Legendre symbol \(\left(\frac{a}{p}\right)\) is defined by
\[\left(\frac{a}{p}\right)=\left\{\begin{array}{lcr} \ 1 &\mbox{if a is a quadratic residue of p} \\ \ -1 &\mbox{if a is a quadratic nonresidue of p}. \\ \end{array}\right .\]
Notice that using the previous example, we see that
\[\begin{aligned} && \left(\frac{1}{7}\right)=\left(\frac{2}{7}\right)=\left(\frac{4}{7}\right)=1\\ && \left(\frac{3}{7}\right)=\left(\frac{5}{7}\right)=\left(\frac{6}{7}\right)=-1\end{aligned}\]
In the following theorem, we present a way to determine wether an integer is a quadratic residue of a prime.
Euler’s Criterion
Let \(p\neq 2\) be a prime and let \(a\) be a positive integer such that \(p\nmid a\). Then
\[\left(\frac{a}{p}\right)\equiv a^{\phi(p)/2}(mod \ p).\]
Assume that \(\left(\frac{a}{p}\right)=1\). Then the congruence \(x^2\equiv a(mod \ p)\) has a solution say \(x=x'\). According to Fermat’s theorem, we see that
\[a^{\phi(p)/2}=((x')^2)^{\phi(p)/2}\equiv 1(mod\ p).\]
Now if \(\left(\frac{a}{p}\right)=-1\), then \(x^2\equiv a(mod \ p)\) is not solvable. Thus by Theorem 26, we have that for each integer k with \((k,p)=1\) there is an integer \(l\) such that \(kl\equiv a(mod \ p)\). Notice that \(i\neq j\) since \(x^2\equiv a(mod \ p)\) has no solutions. Thus we can couple the integers \(1,2,...,p-1\) into \((p-1)/2\) pairs, each has product \(a\). Multiplying these pairs together, we find out that
\[(p-1)!\equiv a^{\phi(p)/2}(mod \ p).\]
Using Wilson’s Theorem , we get \[\left(\frac{a}{p}\right)=-1\equiv a^{(p-1)/2}(mod \ p).\]
Let \(p=13\) and \(a=3\). Then \(\left(\frac{3}{13}\right)=-1\equiv 3^{6}(mod \ 13).\)
We now prove some properties of Legendre symbol.
Let \(p\neq 2\) be a prime. Let \(a\) and \(b\) be integers such that \(p\nmid a\), \(p\nmid b\) and \(p\mid (a-b)\) then \[\left(\frac{a}{p}\right)=\left(\frac{b}{p}\right).\]
Since \(p\mid (a-b)\), then \(x^2\equiv a(mod \ p)\) has a solution if and only if \(x^2\equiv b(mod \ p)\) has a solution. Hence \[\left(\frac{a}{p}\right)=\left(\frac{b}{p}\right)\]
Let \(p\neq 2\) be a prime. Let \(a\) and \(b\) be integers such that \(p\nmid a\), \(p\nmid b\) then \[\left(\frac{a}{p}\right)\left(\frac{b}{p}\right)=\left(\frac{ab}{p}\right)\]
By Euler’s criterion, we have \[\left(\frac{a}{p}\right)\equiv a^{\phi(p)/2}(mod \ p)\] and \[\left(\frac{b}{p}\right)\equiv b^{\phi(p)/2}(mod \ p).\] Thus we get \[\left(\frac{a}{p}\right)\left(\frac{b}{p}\right) \equiv (ab)^{\phi(p)/2}\equiv \left(\frac{ab}{p}\right)(mod \ p).\] We now show when is \(-1\) a quadratic residue of a prime \(p\).
If \(p\neq 2\) is a, then \[\left(\frac{-1}{p}\right)=\left\{\begin{array}{lcr} \ 1 &{\mbox{if}\ p\equiv 1(mod \ 4)} \\ \ -1 &{\mbox{if}\ p\equiv -1(mod \ 4)}. \\ \end{array}\right .\]
By Euler’s criterion, we know that \[\left(\frac{a}{p}\right)=(-1)^{\phi(p)/2}(mod \ p)\] If \(4\mid (p-1)\), then \(p=4m+1\) for some integer \(m\) and thus we get \[(-1)^{\phi(p)/2}=(-1)^{2m}=1.\] and if \(4\mid (p-3)\), then \(p=4m+3\) for some integer \(m\) and we also get \[(-1)^{\phi(p)/2}=(-1)^{2m+1}=-1.\]
We now determine when \(2\) is a quadratic residue of a prime \(p\).
For every odd prime \(p\) we have
\[\left(\frac{2}{p}\right)=\left\{\begin{array}{lcr} \ 1 &{\mbox{if}\ p\equiv \pm1(mod \ 8)} \\ \ -1 &{\mbox{if}\ p\equiv \pm 3(mod \ 8)}. \\ \end{array}\right .\]
Consider the following \((p-1)/2\) congruences
\[\begin{aligned} p-1&\equiv& 1(-1)^1 \ \ \ (mod \ p)\\ 2&\equiv& 2(-1)^2 \ \ \ (mod \ p)\\ p-3&\equiv& 3(-1)^3 \ \ \ (mod \ p)\\ 4&\equiv& 4(-1)^4 \ \ \ (mod \ p)\\ &.& \\ &.& \\ &.& \\ r&\equiv& \frac{p-1}{2}(-1)^{(p-1)/2} \ \ \ (mod \ p),\\\end{aligned}\]
where \(r\) is either \(p-(p-1)/2\) or \((p-1)/2\). Multiplying all these equations we get,
\[2.4.6...(p-1)\equiv \left(\frac{p-1}{2}\right)!(-1)^{1+2+...+(p-1)/2} \ \ \ (mod \ p).\]
This gives us
\[2^{(p-1)/2}\left(\frac{p-1}{2}\right)! \equiv \left(\frac{p-1}{2}\right)!(-1)^{(p^2-1)/8} (mod \ p).\]
Now notice that \(\left(\frac{p-1}{2}\right)!\not\equiv 0(mod \ p)\) and thus we get
\[2^{(p-1)/2}\equiv (-1)^{(p^2-1)/8}(mod \ p).\]
Note also that by Euler’s criterion, we get
\[2^{\phi(p)/2}\equiv \left(\frac{2}{p}\right)(mod \ p),\]
and since each member is 1 or -1 the two members are equal.
We now present an important lemma that determines whether an integer is a quadratic residue of a prime or not.
Gauss’s Lemma
Let \(p\neq 2\) be a prime and \(a\) a relatively prime integer to \(p\). If \(k\) counts the number of least positive residues of the integers \(a, 2a,...,((p-1)/2)a\) that are greater than \(p/2\), then
\[\left(\frac{a}{p}\right)=(-1)^k.\]
Let \(m_1,m_2,...,m_s\) be those integers greater than \(p/2\) in the set of the least positive residues of the integers \(a, 2a,...,((p-1)/2)a\) and let \(n_1,n_2,...,n_t\) be those less than \(p/2\). We now show that
\[p-m_1,p-m_2,...,p-m_k,p-n_1,p-n_2,...,p-n_t\]
are precisely the integers \[1,2,...,(p-1)/2,\] in the same order.
So we shall show that no two integers of these are congruent modulo \(p\), because there are exactly \((p-1)/2\) numbers in the set, and all are positive integers less than or equal to \((p-1)/2\). Notice that \(m_i\not\equiv m_j (\mod \ p)\) for all \(i\neq j\) and \(n_i\not\equiv n_j (\mod \ p)\) for all \(i\neq j\). If any of these congruences fail, then we will have that \(r\equiv s(mod \ p)\) assuming that \(ra\equiv sa(mod \ p)\). Also any of the integers \(p-m_i\) can be congruent to any of the \(n_i\)’s. Because if such congruence holds, then we have \(ra\equiv p-sa(mod \ p)\), so that \(ra\equiv -sa(mod \ p)\). Because \(p\nmid a\), this implies that \(r\equiv -s(mod \ p)\), which is impossible. We conclude that
\[\prod_{i=1}^k(p-m_i)\prod_{i=1}^tn_i\equiv \left(\frac{p-1}{2}\right)!(mod \ p),\]
which implies
\[(-1)^sm_1m_2...(p-m_k)n_1n_2...n_t\equiv \left(\frac{p-1}{2}\right)!(mod \ p),\]
Simplifying, we get
\[m_1m_2...(p-m_k)n_1n_2...n_t\equiv a.2a...((p-1)/2)= a^{(p-1)/2}((p-1)/2)!( mod \ p).\]
As a result, we have that
\[a^{(p-1)/2}((p-1)/2)!\equiv ((p-1)/2)!(mod \ p)\]
Note that since \((p,((p-1)/2)!)=1\), we get
\[(-1)^ka^{(p-1)/2}\equiv 1(mod \ p).\]
Thus we get
\[a^{(p-1)/2}\equiv(-1)^k(mod \ p).\]
Using Euler’s criterion, the result follows.
To find \(\left(\frac{5}{13}\right)\) using Gauss’s lemma, we calculate
\[\sum_{i=1}^6[5i/13]=[5/13]+[10/13]+[15/13]+[20/13]+[25/13]+[30/13]=5\] Thus we get \(\left(\frac{5}{13}\right)=(-1)^5=-1\).
Exercises
- Find all quadratic residues of 3
- Find all quadratic residues of 19.
- Find the value of Legendre symbol \(\left(\frac{j}{7}\right)\) for \(j=1,2,3,4,5,6\).
- Evaluate the Legendre symbol \(\left(\frac{7}{11}\right)\) by using Euler’s criterion.
- Let \(a\) and \(b\) be integers not divisible by \(p\). Show that either one or all three of the integers \(a,b\) and \(ab\) are quadratic residues of \(p\).
- Let \(p\) be a prime and \(a\) be a quadratic residue of \(p\). Show that if \(p\equiv 1(mod \ 4)\), then \(-a\) is also a quadratic residue of \(p\), whereas if \(p\equiv 3(mod \ 4)\), then \(-a\) is a quadratic nonresidue of \(p\).
- Show that if \(p\) is an odd prime and a is an integer not divisible by \(p\) then \(\left(\frac{a^2}{p}\right)=1\).
Contributors and Attributions
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Dr. Wissam Raji, Ph.D., of the American University in Beirut. His work was selected by the Saylor Foundation’s Open Textbook Challenge for public release under a Creative Commons Attribution ( CC BY ) license.