document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1207684>Question 1207684</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find all integers $n$, $0 \le n < 163$, such that $n$ is its own inverse modulo $8.$ </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #845715 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=845715\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Tutor ikleyn has pointed out that n^2 = 1 (mod 8) has solutions: <BR>\n" );
document.write( "n = 1, n = 3, n = 5, n = 7\r<BR>\n" );
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document.write( "The interesting thing is that the n = 1 and n = 7 cases pair up.<BR>\n" );
document.write( "Notice how n = 7 is one short of 8, so we can think of it like n = -1 in mod 8.<BR>\n" );
document.write( "Squaring -1 will generate +1 or simply 1. \r<BR>\n" );
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document.write( "Similarly, the n = 3 and n = 5 cases pair up together. <BR>\n" );
document.write( "Notice n = 5 is three short of 8, in which we can think of it like n = -3 (mod 8). Squaring this leads to +9 = 9 = 1 (mod 8).\r<BR>\n" );
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document.write( "Slight tangent aside, the four classes of solutions were<BR>\n" );
document.write( "n = 1, n = 3, n = 5, n = 7\r<BR>\n" );
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document.write( "Your teacher is asking you to look through the interval <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0+%3C=+n+%3C+163 ALIGN=MIDDLE  ALT=\"0+%3C=+n+%3C+163\"   DISABLED_style= \"max-width:90%; height:auto;\" > to see which of those values of n work or not.\r<BR>\n" );
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document.write( "Let's focus on the n = 1 case.<BR>\n" );
document.write( "This is when the numbers are of the form 8k+1, since each produces remainder 1 in mod 8.<BR>\n" );
document.write( "Look at a few values of k.<BR>\n" );
document.write( "n = 8k+1 = 8*0+1 = 1<BR>\n" );
document.write( "n = 8k+1 = 8*1+1 = 9<BR>\n" );
document.write( "n = 8k+1 = 8*2+1 = 17<BR>\n" );
document.write( "n = 8k+1 = 8*3+1 = 25<BR>\n" );
document.write( "Another way to generate this sequence is to start at 1, and add 8 to each term. <BR>\n" );
document.write( "The question is now: when does this subsequence stop?<BR>\n" );
document.write( "Let's set it equal to 163 and see what happens.<BR>\n" );
document.write( "8k+1 = 163<BR>\n" );
document.write( "8k = 163-1<BR>\n" );
document.write( "8k = 162<BR>\n" );
document.write( "k = 162/8<BR>\n" );
document.write( "k = 20.25<BR>\n" );
document.write( "If k = 20 then 8k+1 = 8*20+1 = 161<BR>\n" );
document.write( "If k = 21 then 8k+1 = 8*21+1 = 167<BR>\n" );
document.write( "Therefore the highest we can go in this congruence class is 161.\r<BR>\n" );
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document.write( "All of these values<BR>\n" );
document.write( "{1, 9, 17, ..., 153, 161}<BR>\n" );
document.write( "fit the n = 1 congruence class and are a subset of the overall solution space.<BR>\n" );
document.write( "When dividing each of those items over 8, we'll get some quotient with remainder 1.\r<BR>\n" );
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document.write( "You will follow similar ideas for n = 3, n = 5, and n = 7. <BR>\n" );
document.write( "I'll let the student determine those. <BR>\n" );
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