document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209952>Question 1209952</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The function f(n) is defined for all integers n, such that\r<BR>\n" );
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document.write( "f(x) + f(y) = f(x + y) - 4xy - 1 + f(x^2) + f(y^2)<BR>\n" );
document.write( "for all integers x and y, and f(1) = 1.  Find f(n).<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851203 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=851203\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> We are given the functional equation:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "f(x) + f(y) = f(x + y) - 4xy - 1 + f(x^2) + f(y^2)<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "for all integers \( x, y \), along with the initial condition \( f(1) = 1 \). Our goal is to determine \( f(n) \) explicitly.\r<BR>\n" );
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document.write( "### Step 1: Plugging in \( x = 0, y = 0 \)\r<BR>\n" );
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document.write( "Substituting \( x = 0, y = 0 \) into the given equation:\r<BR>\n" );
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document.write( "f(0) + f(0) = f(0) - 4(0)(0) - 1 + f(0^2) + f(0^2)<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Since \( f(0^2) = f(0) \), we simplify:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "2f(0) = f(0) - 1 + 2f(0)<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Canceling \( 2f(0) \) from both sides, we get:\r<BR>\n" );
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document.write( "0 = -1 \Rightarrow f(0) = 0.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "### Step 2: Plugging in \( y = 1 \)\r<BR>\n" );
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document.write( "Setting \( y = 1 \) in the functional equation:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "f(x) + f(1) = f(x + 1) - 4x(1) - 1 + f(x^2) + f(1^2).<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Since \( f(1) = 1 \) and \( f(1^2) = f(1) = 1 \), we substitute:\r<BR>\n" );
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document.write( "f(x) + 1 = f(x + 1) - 4x - 1 + f(x^2) + 1.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Simplify:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "f(x) = f(x + 1) - 4x + f(x^2).<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "### Step 3: Plugging in \( x = 1, y = 1 \)\r<BR>\n" );
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document.write( "Setting \( x = y = 1 \):\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "f(1) + f(1) = f(2) - 4(1)(1) - 1 + f(1^2) + f(1^2).<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Using \( f(1) = 1 \):\r<BR>\n" );
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document.write( "1 + 1 = f(2) - 4 - 1 + 1 + 1.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "2 = f(2) - 3.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "f(2) = 5.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "### Step 4: Plugging in \( x = 2, y = 1 \)\r<BR>\n" );
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document.write( "f(2) + f(1) = f(3) - 4(2)(1) - 1 + f(2^2) + f(1^2).<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Using \( f(2) = 5 \), \( f(1) = 1 \), \( f(1^2) = 1 \):\r<BR>\n" );
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document.write( "5 + 1 = f(3) - 8 - 1 + f(4) + 1.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "6 = f(3) - 8 + f(4) + 1.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "5 = f(3) - 8 + f(4).<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Rearrange:\r<BR>\n" );
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document.write( "f(3) + f(4) = 13.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "### Step 5: Identifying a Pattern\r<BR>\n" );
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document.write( "Based on observed values \( f(1) = 1 \), \( f(2) = 5 \), and the recurrence relation, we suspect \( f(n) \) follows a quadratic pattern. Assume:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "f(n) = an^2 + bn + c.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Using \( f(1) = 1 \):\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "a(1)^2 + b(1) + c = 1 \Rightarrow a + b + c = 1.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Using \( f(2) = 5 \):\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "a(2)^2 + b(2) + c = 5 \Rightarrow 4a + 2b + c = 5.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Using \( f(0) = 0 \):\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "a(0)^2 + b(0) + c = 0 \Rightarrow c = 0.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "So,\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "a + b = 1, \quad 4a + 2b = 5.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Solving for \( a, b \):\r<BR>\n" );
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document.write( "b = 1 - a,<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "4a + 2(1 - a) = 5.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "4a + 2 - 2a = 5.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "2a = 3.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "a = \frac{3}{2}, \quad b = 1 - \frac{3}{2} = -\frac{1}{2}.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Thus, \r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "f(n) = \frac{3}{2} n^2 - \frac{1}{2} n.<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Verifying,\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "f(n) = \frac{3}{2} n(n - 1).<BR>\n" );
document.write( "\]\r<BR>\n" );
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document.write( "Final Answer:\r<BR>\n" );
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document.write( "\[<BR>\n" );
document.write( "\boxed{f(n) = \frac{3}{2} n(n-1).}<BR>\n" );
document.write( "\] </SPAN>\n" );
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