document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=549954>Question 549954</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The equation x3 +3xy + y3 = 1 is solved in integers. Find the possible values of x-y </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #358530 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=358530\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> From the binomial expansion, we have\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3\">, which we can write as\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE (x+y)^3 = x^3 + 3xy(x+y) + y^3\">.\r<BR>\n" );
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document.write( "Clearly, all (x,y) satisfying x+y = 1 work. Note that the other tutor also stated x+y = 1, using a valid argument, but the solution is much longer.\r<BR>\n" );
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document.write( "Now suppose x+y was not equal to 1. We may treat the original equation as a cubic equation by fixing a value for y and solving for x. In other words, we want to find integer roots for the cubic equation\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE x^3 + 3xy + (y^3 - 1) = 0\"> other than x = 1-y.\r<BR>\n" );
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document.write( "Here, we use the fact that if r is a root of a polynomial, then 1-r is a factor. We divide both sides by x - (1-y) or x+y-1:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE x^2 + x(1-y) + (y^2 + y + 1) = 0\">. Solve for x via the quadratic formula.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE x = \frac{y-1 \pm \sqrt{(y^2 - 2y + 1) - 4(y^2 + y + 1)}}{2}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE = \frac{y-1 \pm \sqrt{-3y^2 - 6y - 3}}{2}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE = \frac{y-1 \pm \sqrt{-3(y^2 + 2y + 1)}}{2}\">\r<BR>\n" );
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document.write( "One thing to note: the expression inside the radical is equal to -3(y+1)^2. Since the square of a number is always non-negative, we will never have a real root, unless the expression inside the radical is equal to zero. Hence, we want y to equal -1. If y = -1, then x is also equal to -1. It is seen that (x,y) = (-1,-1) satisfies the original equation.\r<BR>\n" );
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document.write( "Therefore, the only solutions to the original equation are ordered pairs (x,y) satisfying x+y = 1 or (-1,-1). We want to evaluate x-y, which is equal to (x+y)-2y, or 1-2y, which can be any odd number (since y can equal anything). Also, (-1,-1) yields x-y = 0. Hence, the possible values for x-y are odd numbers and 0. </SPAN>\n" );
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