document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1059605>Question 1059605</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Prove that if 1 is added to the product of any four <BR>\n" );
document.write( "consecutive integers, the sum is a perfect square.  <BR>\n" );
document.write( "Thank you in advance. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #674658 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=674658\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x(x+1)(x+2)(x+3)+1=x(x+3)(x+2)(x+1)+1<BR>\n" );
document.write( "so (x^2+3x)(x^2+3x+2)+1=(x^2+3x+1)(x^2+3x+2)-(x^2+3x+2)+1<BR>\n" );
document.write( "You are converting the first term to (x^2+3x+1), and that is adding x^2+3x+2 to that side.  That means you have to subtract it.<BR>\n" );
document.write( "Convert the (x^2+3x+2) to (x^2+3x+1). But that requires adding the factor x^2+3x+1. because by subtracting one from that term, which is multiplied by (x^2+3x+1), you have to add it back.<BR>\n" );
document.write( "Then you have (x^2+3x+1)(x^2+3x+1)-(x^2+3x+2)+(x^2+3x+1)+1.  The last two terms disappear because distributing the minus sign give -x^2-3x-2+x^2+3x+1+1.<BR>\n" );
document.write( "You are left with (x^2+3x+1)^2, which is a perfect square.<BR>\n" );
document.write( "Pick an x, like 7<BR>\n" );
document.write( "7,8,9,10 has a product of 5040 and 5041, 1 more, is a perfect square of <BR>\n" );
document.write( "x^2+3x+1, or 49+21+1, or 71.\r<BR>\n" );
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