document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1177009>Question 1177009</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Prove that 5^{3^n} + 1 is divisible by 3^{n + 1} for all nonnegative integers n. Prove this using induction. Can you please explain in detail? Thank you so much! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850539 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=850539\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Absolutely, let's prove that $5^{3^n} + 1$ is divisible by $3^{n+1}$ for all nonnegative integers $n$ using the Lifting The Exponent Lemma (LTE).\r<BR>\n" );
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document.write( "**Understanding the Problem**\r<BR>\n" );
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document.write( "We want to show that $3^{n+1} \mid (5^{3^n} + 1)$ for all $n \geq 0$.\r<BR>\n" );
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document.write( "**Lifting The Exponent Lemma (LTE)**\r<BR>\n" );
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document.write( "The LTE Lemma states that if $x$ and $y$ are integers, $n$ is a positive integer, and $p$ is an odd prime such that $p \mid (x+y)$ but $p \nmid x$ and $p \nmid y$, then:\r<BR>\n" );
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document.write( "$v_p(x^n + y^n) = v_p(x+y) + v_p(n)$ if $n$ is odd.\r<BR>\n" );
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document.write( "Here, $v_p(m)$ denotes the highest power of $p$ that divides $m$.\r<BR>\n" );
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document.write( "**Applying LTE**\r<BR>\n" );
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document.write( "1. **Base Case (n = 0):**<BR>\n" );
document.write( "   When $n=0$, we have $5^{3^0} + 1 = 5^1 + 1 = 6$. We need to show that $3^{0+1} = 3^1 = 3$ divides $6$. Clearly, $3 \mid 6$, so the base case holds.\r<BR>\n" );
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document.write( "2. **Inductive Step:**<BR>\n" );
document.write( "   Let's assume that $3^{n+1} \mid (5^{3^n} + 1)$ for some nonnegative integer $n$. We want to show that $3^{n+2} \mid (5^{3^{n+1}} + 1)$.\r<BR>\n" );
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document.write( "   We have $5^{3^{n+1}} + 1 = (5^{3^n})^3 + 1^3$.<BR>\n" );
document.write( "   Let $x = 5^{3^n}$ and $y = 1$. Then we have $x^3 + y^3$.\r<BR>\n" );
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document.write( "   We know that $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$.<BR>\n" );
document.write( "   So, $5^{3^{n+1}} + 1 = (5^{3^n} + 1)((5^{3^n})^2 - 5^{3^n} + 1)$.\r<BR>\n" );
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document.write( "   We want to find $v_3(5^{3^{n+1}} + 1)$.<BR>\n" );
document.write( "   We can apply LTE to $x^3 + y^3$:\r<BR>\n" );
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document.write( "   $v_3(5^{3^{n+1}} + 1) = v_3((5^{3^n})^3 + 1^3) = v_3(5^{3^n} + 1) + v_3(3)$.\r<BR>\n" );
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document.write( "   By the inductive hypothesis, $3^{n+1} \mid (5^{3^n} + 1)$. Therefore, $v_3(5^{3^n} + 1) = n+1$.\r<BR>\n" );
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document.write( "   Also, $v_3(3) = 1$.\r<BR>\n" );
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document.write( "   Thus, $v_3(5^{3^{n+1}} + 1) = (n+1) + 1 = n+2$.\r<BR>\n" );
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document.write( "   This means that $3^{n+2} \mid (5^{3^{n+1}} + 1)$.\r<BR>\n" );
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document.write( "**Conclusion**\r<BR>\n" );
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document.write( "By the principle of mathematical induction, $5^{3^n} + 1$ is divisible by $3^{n+1}$ for all nonnegative integers $n$.\r<BR>\n" );
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document.write( "**Explanation of LTE Application**\r<BR>\n" );
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document.write( "* We identified $x = 5^{3^n}$ and $y = 1$.<BR>\n" );
document.write( "* We observed that $3 \mid (5^{3^n} + 1)$ by the inductive hypothesis.<BR>\n" );
document.write( "* We noted that $3 \nmid 5^{3^n}$ and $3 \nmid 1$.<BR>\n" );
document.write( "* We applied the LTE formula for odd $n=3$, which gave us $v_3(x^3 + y^3) = v_3(x+y) + v_3(3)$.<BR>\n" );
document.write( "* We used the inductive assumption to replace $v_3(5^{3^n} + 1)$ with $n+1$.<BR>\n" );
document.write( "* We used the fact that $v_3(3) = 1$.\r<BR>\n" );
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document.write( "This allowed us to conclude that $v_3(5^{3^{n+1}} + 1) = n+2$, which implies $3^{n+2} \mid (5^{3^{n+1}} + 1)$.<BR>\n" );
document.write( " </SPAN>\n" );
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