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Difference between revisions of "2007 USAMO Problems/Problem 5"

 
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== Problem ==
 
== Problem ==
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(''Titu Andreescu'') Prove that for every [[nonnegative]] [[integer]] <math>n</math>, the number <math>7^{7^n}+1</math> is the [[product]] of at least <math>2n+3</math> (not necessarily distinct) [[prime]]s.
  
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== Solution ==
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==Solutions==
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=== Solution 1 ===
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The proof is by induction. The base is provided by the <math>n = 0</math> case, where <math>7^{7^0} + 1 = 7^1 + 1 = 2^3</math>. To prove the inductive step, it suffices to show that if <math>x = 7^{2m - 1}</math> for some positive integer <math>m</math> then <math>(x^7 + 1)/(x + 1)</math> is composite. As a consequence, <math>x^7 + 1</math> has at least two more prime factors than does <math>x + 1</math>. To confirm that <math>(x^7 + 1)/(x + 1)</math> is composite, observe that
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<cmath>\begin{align*}
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\frac{x^7 + 1}{x + 1} &= \frac{(x + 1)^7 - ((x + 1)^7 - (x^7 + 1))}{x + 1} \\
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&= (x + 1)^6 - \frac{7x(x^5 + 3x^4 + 5x^3 + 5x^2 + 3x + 1)}{x + 1} \\
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&= (x + 1)^6 - 7x(x^4 + 2x^3 + 3x^2 + 2x + 1) \\
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&= (x + 1)^6 - 7^{2m}(x^2 + x + 1)^2 \\
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&= \{(x + 1)^3 - 7^m(x^2 + x + 1)\}\{(x + 1)^3 + 7^m(x^2 + x + 1)\}.
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\end{align*}</cmath>
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Also each factor exceeds 1. It suffices to check the smaller one; <math>\sqrt{7x}\leq x</math> gives
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<cmath>\begin{align*}
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(x + 1)^3 - 7^m(x^2 + x + 1) &= (x + 1)^3 - \sqrt{7x}(x^2 + x + 1) \\
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&\geq x^3 + 3x^2 + 3x + 1 - x(x^2 + x + 1) \\
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&= 2x^2 + 2x + 1\geq 113 > 1.
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\end{align*}</cmath>
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Hence <math>(x^7 + 1)/(x + 1)</math> is composite and the proof is complete.
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{{alternate solutions}}
  
 
== See also ==
 
== See also ==
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* <url>viewtopic.php?t=145849 Discussion on AoPS/MathLinks</url>
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{{USAMO newbox|year=2007|num-b=4|num-a=6}}
 
{{USAMO newbox|year=2007|num-b=4|num-a=6}}
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[[Category:Olympiad Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 08:47, 7 August 2014

Problem

(Titu Andreescu) Prove that for every nonnegative integer $n$, the number $7^{7^n}+1$ is the product of at least $2n+3$ (not necessarily distinct) primes.

Solutions

Solution 1

The proof is by induction. The base is provided by the $n = 0$ case, where $7^{7^0} + 1 = 7^1 + 1 = 2^3$. To prove the inductive step, it suffices to show that if $x = 7^{2m - 1}$ for some positive integer $m$ then $(x^7 + 1)/(x + 1)$ is composite. As a consequence, $x^7 + 1$ has at least two more prime factors than does $x + 1$. To confirm that $(x^7 + 1)/(x + 1)$ is composite, observe that \begin{align*} \frac{x^7 + 1}{x + 1} &= \frac{(x + 1)^7 - ((x + 1)^7 - (x^7 + 1))}{x + 1} \\ &= (x + 1)^6 - \frac{7x(x^5 + 3x^4 + 5x^3 + 5x^2 + 3x + 1)}{x + 1} \\ &= (x + 1)^6 - 7x(x^4 + 2x^3 + 3x^2 + 2x + 1) \\ &= (x + 1)^6 - 7^{2m}(x^2 + x + 1)^2 \\ &= \{(x + 1)^3 - 7^m(x^2 + x + 1)\}\{(x + 1)^3 + 7^m(x^2 + x + 1)\}. \end{align*} Also each factor exceeds 1. It suffices to check the smaller one; $\sqrt{7x}\leq x$ gives \begin{align*} (x + 1)^3 - 7^m(x^2 + x + 1) &= (x + 1)^3 - \sqrt{7x}(x^2 + x + 1) \\ &\geq x^3 + 3x^2 + 3x + 1 - x(x^2 + x + 1) \\ &= 2x^2 + 2x + 1\geq 113 > 1. \end{align*} Hence $(x^7 + 1)/(x + 1)$ is composite and the proof is complete.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

  • <url>viewtopic.php?t=145849 Discussion on AoPS/MathLinks</url>
2007 USAMO (Problems • Resources)
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Problem 6
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All USAMO Problems and Solutions

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