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Mathematics LibreTexts

5.1: The order of Integers and Primitive Roots

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    In this section, we study the order of an integer modulo \(n\), where \(n\) is positive. We also define primitive roots and related results. Euler’s theorem in Chapter 4 states that if a positive integer \(a\) is relatively prime to \(n\), then \(a^{\phi(n)}\equiv 1 (mod \ n)\). Thus by the well ordering principle, there is a least positive integer \(x\) that satisfies this congruence \(a^{x}\equiv 1 (mod \ n)\).

    Let \((a,b)=1\). The smallest positive integer \(x\) such that \(a^{x} \equiv 1(mod \ b)\) is called the order of \(a\) modulo \(b\). We denote the order of \(a\) modulo \(b\) by \(ord_ba\).

    \(ord_72=3\) since \(2^3\equiv 1(mod \ 7)\) while \(2^1\equiv 2(mod \ 7)\) and \(2^2\equiv 4(mod \ 7)\).

    To find all integers \(x\) such that \(a^x\equiv 1(mod \ b)\), we need the following theorem.

    If \((a,b)=1\) with \(b>0\), then the positive integer \(x\) is a solution of the congruence \(a^x\equiv 1(mod \ b)\) if and only if \(ord_ba\mid x\).

    Having \(ord_ba\mid x\), then we have that \(x=k.ord_ba\) for some positive integer \(k\). Thus \[a^x=a^{kord_ba}=(a^{ord_ba})^k\equiv 1(mod \ b).\] Now if \(a^x\equiv 1(mod \ b)\), we use the division algorithm to write \[x=q ord_ba+r, \ \ \ 0\leq r<ord_ba.\] Thus we see that \[a^x\equiv a^{qord_ba+r}\equiv (a^{ord_ba})^qa^r\equiv a^r (mod \ b).\] Now since \(a^x\equiv 1(mod \ b)\),we have \(a^r\equiv 1(mod \ b)\). Since \(ord_ba\), we get \(r=0\). Thus \(x=q.ord_ba\) and hence \(ord_ba\mid x\).

    Since \(ord_72=3\), then \(2^{15}\equiv 1(mod \ 7)\) while 10 is not a solution for \(2^x\equiv 1 (mod \ 7)\).

    If \((a,b)=1\) with \(b>0\), then \[a^i\equiv a^j(mod \ b)\] where \(i\) and \(j\) are nonnegative integers, if and only if \[i\equiv j(mod \ ord_ba)\]

    Suppose that \[i\equiv j(mod \ ord_ba)\ \ \mbox{and}\ \ 0\leq j\leq i.\] Then we have \(i-j=k.ord_ba\), where \(k\) is a positive integer. Hence \[a^i=a^{j+k.ord_ba}=a^j(a^{ord_ba})^k\equiv a^j (mod \ b).\] Assume now that \(a^i\equiv a^j(mod \ b)\) with \(i\geq j\). Thus we have \[a^i\equiv a^ja^{i-j}\equiv a^j(mod \ b)\] Since \((a,b)=1\), we have \((a^j,b)=1\) and thus by Theorem 22, we get \[a^{i-j}\equiv 1(mod \ b).\] By theorem 54, we get that \(ord_ba \mid (i-j)\) and hence \(i\equiv j(mod \ b)\).

    We introduce now primitive roots and discuss their properties. We are interested in integers whose order modulo another integer is \(\phi(b)\). In one of the exercises, one is asked to prove that if \(a\)and \(b\) are relatively prime then \(ord_ba \mid \phi(b)\).

    If \((r,m)=1\) with \(m>0\) and if \(ord_mr=\phi(m)\) then \(r\) is called a primitive root modulo \(m\).

    Notice that \(\phi(7)=6\) hence \(2\) is not a primitive root modulo \(7\). While \(ord_73=6\) and thus \(3\) is a primitive root modulo \(7\).

    If \((r,m)=1\) with \(m>0\) and if \(r\) is a primitive root modulo \(n\), then the integers \(\{r^1,r^2,...r^{\phi(m)}\}\) form a reduced residue set modulo \(m\).

    To prove that the set \(\{r^1,r^2,...r^{\phi(m)}\}\) form a reduced residue set modulo \(m\) we need to show that every two of them are relatively prime and that no two of them are congruent modulo \(m\). Since \((r,m)=1\), it follows that \((r^n,m)=1\) for all positive integers \(n\). Hence all the powers of \(r\) are relatively prime to \(m\). To show that no two powers in the above set are equivalent modulo \(m\), assume that \[r^i\equiv r^j(mod \ m).\] By Theorem 55, we see that \[i\equiv j(mod \ ord_m\phi(m)).\] Notice that \(1\leq i,j\leq \phi(m)\) and hence \(i=j\).

    If \(ord_ma=t\) and if \(u\) is a positive integer, then \[ord_m(a^u)=t/(t,u).\]

    Let \[v=ord_m(a^u),\ \ w=(t,u), \ \ t=t_1w \mbox{and}\ \ u=u_1w.\] Notice that \((t_1,u_1)=1.\)

    Because \(t_1=t/(t,u)\), we want to show that \(ord_m(a^u)=t_1\). To do this, we will show that \((a^{u})^{t_1}\equiv 1(mod \ m)\) and that if \((a^u)^v\equiv 1(mod \ m)\), then \(t_1\mid v\). First note that \[(a^u)^{t_1}=(a^{u_1w})^{(t/w)}=(a^t)^{u_1}\equiv 1(mod \ m).\] Hence by Theorem 54, we have \(v\mid t_1\). Now on the other hand, since \[(a^u)^v=a^{uv}\equiv 1(mod \ m),\] we know that \(t\mid uv\). Hence \(t_1w\mid u_1wv\) and hence \(t_1\mid u_1v\). Because \((t_1,u_1)=1\), we see that \(t_1\mid v\). Since \(v\mid t_1\) and \(t_1\mid v\), we conclude that \(v=t_1=t/w=t/(t,u)\).

    We see that \(ord_73^4=6/(6,4)\) since \(ord_73=6\).

    Let \(r\) be a primitive root modulo \(m\), where \(m\) is a positive integer, \(m>1\). Then \(r^u\) is a primitive root modulo \(m\) if and only if \((u,\phi(m))=1\).

    By Theorem 57, we see that \[ord_mr^u=ord_mr/(u,ord_mr)=\phi(m)/(u,\phi(m)).\] Thus \(ord_mr^u=\phi(m)\) and \(r^u\) is a primitive root if and only if \((u,\phi(m))=1\).

    The above corollary leads to the following theorem

    If the positive integer \(m\) has a primitive root, then it has a total of \(\phi(\phi(m))\) incongruent primitive roots.

    Let \(r\) be a primitive root modulo \(m\). By Theorem 56, we see that \(\{r^1,r^2,...,r^{\phi(m)}\}\) form a reduced residue system modulo \(n\). By Corollary 1, it is known that \(r^u\) is a primitive root modulo \(m\) if and only if \((u,\phi(m))=1\). Thus we have exactly \(\phi(\phi(m))\) such integers \(u\) that are relatively prime to \(\phi(m)\) and hence there are exactly \(\phi(\phi(m))\) primitive roots modulo \(m\).


    1. Determine \(ord_{13}10\).

    2. Determine \(ord_{11}3.\)

    3. Show that 5 is a primitive root of 6.

    4. Show that if \(\bar{a}\) is an inverse of \(a\) modulo \(n\), then \(ord_na=ord_n\bar{a}\).

    5. Show that if \(n\) is a positive integer, and \(a\) and \(b\) are integers relatively prime to \(n\) such that \((ord_na,ord_nb)=1\), then \(ord_n(ab)=ord_na.ord_nb\).

    6. Show that if \(a\) is an integer relatively prime to the positive integer \(m\) and \(ord_ma=st\), then \(ord_ma^t=s\).

    7. Show that if \(a\) and \(n\) are relatively prime with \(n>0\), then \(ord_na \mid \phi(n)\).


    • 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.