document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=507684>Question 507684</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Conjecture: Find all positive integers 'a' such that there exists an integer 'm' with the property that a|(m^2+1) and a|((m+1)^2 +1).\r<BR>\n" );
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document.write( "Hint: First show that 'a' must also divide 2m+1. \r<BR>\n" );
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document.write( "I know that when m divides n it can be defined as n=(m)(q). So in this case it would be (m^2+1)=(a)(q), and so forth with the other problems. I'm not sure if this is correct but since (m^2+1)=(a)(q) and ((m+1)^2 +1)=(a)(q) so I set <BR>\n" );
document.write( "(m^2+1)=((m+1)^2 +1). I then foiled, moved everything to one side and simplified and got 2m+1 with is relevant to the hint, I think.  I don't know if I'm going in the right direction but I've seem to have hit a wall. Also, I don't really know how to show that a|2m+1. Please help. Thanks.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #340711 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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/cgi-bin/show-face.mpl?user=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=340711\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You somehow set m^2 + 1 and (m+1)^2 + 1 equal to each other, which cannot be true.\r<BR>\n" );
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document.write( "We can definitely say that a|(2m+1), but this is a weaker statement. Fortunately we can conclude that a is odd. In addition, we can state that a|(m^2 - 2m), by using a modular arithmetic argument:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE m^2 + 1 \equiv 0 (mod a)\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE 2m + 1 \equiv 0 (mod a) \Rightarrow (m^2 + 1) - (2m + 1) \equiv 0 (mod a)\">\r<BR>\n" );
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document.write( "So this means that a divides 2m+1 and a divides m(m-2). This is actually quite useful, because not many values of a satisfy. We can say that\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE (2m+1)^2 - 4(m(m-2)) \equiv 0 (mod a) \Rightarrow 12m+1 \equiv 0 (mod a)\">\r<BR>\n" );
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document.write( "Since 2m+1 and 12m+1 are congruent mod a, it follows that their difference (10m) is congruent to 0 mod a. This means either a divides m, or a = 5 (remember a cannot equal 2). a = 5 definitely works (try m = 2, 12, ...). If a divides m, then the only value of a is 1 (since a divides 2m+1). Thus a = 1 and a = 5 are the only solutions. \r<BR>\n" );
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