document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=569241>Question 569241</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let a, b,c be positive integers with a>=b>=c such that\r<BR>\n" );
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document.write( "a^2 - b^2 - c^2 +ab=2011<BR>\n" );
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document.write( "a^2 + 3b^2 + 3c^2 -3ab -2ac-2bc=-1997<BR>\n" );
document.write( "what is a? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #367336 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=367336\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> It is generally best to avoid posting this question until the day after the AMC10/12A exams (yes, I took this exam as well). But since it's pretty late in the day, I'll show the solution. Make sure not to publicize it to anyone else.\r<BR>\n" );
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document.write( "Adding both equations (I should have done this myself during the exam...I tried adding three times the first equation to the second equation and got nowhere).\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE 2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14\">\r<BR>\n" );
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document.write( "This factors to\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE (a-b)^2 + (a-c)^2 + (b-c)^2 = 14\">\r<BR>\n" );
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document.write( "Essentially we have three square numbers adding to 14, so the only possibility is {9,4,1}. Since a and c are the furthest apart, we have\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE (a-c)^2 = 9 \Rightarrow a-c = 3\">\r<BR>\n" );
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document.write( "So our three numbers are either {c,c+1,c+3} or {c,c+2,c+3}. Suppose the numbers are {c,c+1,c+3}. Plugging into the first equation we have\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE (c+3)^2 - (c+1)^2 - c^2 + (c+1)(c+3) = 2011 \Rightarrow c = 250\"> (this is actually a linear equation)\r<BR>\n" );
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document.write( "If c = 250, then a = 253 and we are done. For sake of completeness we can try the other case:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE (c+3)^2 - (c+2)^2 - c^2 + (c+2)(c+3) = 2011\">, however if you solved this, c would not be an integer. Therefore a = 253.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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