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

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.

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 == Problem ==
+(''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.
-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.
+==Solutions==
-__TOC__
-==Hint 1 of 3==
-You may be stuck in "factoring" <math>\frac{a^7 + 1}{a + 1} = a^6 - a^5 + a^4 - a^3 + a^2 - a + 1</math> for <math>a = 7^7^n</math>. Keep trying! This is a #5 on a USAMO!
-==Hint 2 of 3==
-Believe it or not, <math>\frac{a^7 + 1}{a + 1}</math> is a difference of squares!
-==Final Hint==
-One of the perfect squares is <math>(x+1)^6</math>.
-==Solution==
 === Solution 1 ===
-We proceed by induction.
+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
+<cmath>\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*}</cmath>
+Also each factor exceeds 1. It suffices to check the smaller one; <math>\sqrt{7x}\leq x</math> gives
+<cmath>\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*}</cmath>
+Hence <math>(x^7 + 1)/(x + 1)</math> is composite and the proof is complete.
-Let <math>{a_{n}}</math> be <math>7^{7^{n}}+1</math>. The result holds for <math>{n=0}</math> because <math>{a_0 = 2^3}</math> is the product of <math>3</math> primes.
+{{alternate solutions}}
-Now we assume the result holds for <math>{n}</math>. Note that <math>{a_{n}}</math> satisfies the [[recursion]]
+== See also ==
+* <url>viewtopic.php?t=145849 Discussion on AoPS/MathLinks</url>
-<div style="text-align:center;"><math>{a_{n+1}= (a_{n}-1)^{7}+1} = a_{n}\left(a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}\right)</math>.</div>
-Since <math>{a_n - 1}</math> is an odd power of <math>{7}</math>, <math>{7(a_n-1)}</math> is a perfect square. Therefore <math>{a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}}</math> is a difference of squares and thus [[composite]], i.e. it is divisible by <math>{2}</math> primes. By assumption, <math>{a_n}</math> is divisible by <math>{2n + 3}</math> primes. Thus <math>{a_{n+1}}</math> is divisible by <math>{2+ (2n + 3) = 2(n+1) + 3}</math> primes as desired.
-=== Solution 2 ===
-Notice that <math>7^{{7}^{k+1}}+1=(7+1) \frac{7^{7}+1}{7+1} \cdot \frac{7^{7^2} + 1}{7^{7^1} + 1} \cdots  \frac{7^{7^{k+1}}+1}{7^{7^{k}}+1}</math>. Therefore it suffices to show that <math>\frac{7^{7^{k+1}}+1}{7^{7^{k}}+1}</math> is composite.
-Let <math>x=7^{7^{k}}</math>. The expression becomes
-<div style="text-align:center;"><math>\frac{7^{7^{k+1}}+1}{7^{7^{k}}+1} = \frac{x^7 + 1}{x + 1}</math></div>
-which is the shortened form of the [[geometric series]] <math>x^{6}-x^{5}+x^{4}-x^3 + x^2 - x+1</math>. This can be [[factor]]ed as <math>(x^{3}+3x^{2}+3x+1)^{2}-7x(x^{2}+x+1)^{2}</math>.
-Since <math>x</math> is an odd power of <math>7</math>, <math>7x</math> is a [[perfect square]], and so we can factor this by difference of squares. Therefore, it is composite.
-== See also ==
 {{USAMO newbox|year=2007|num-b=4|num-a=6}}
-[[Category:Olympiad Algebra Problems]]
+[[Category:Olympiad Number Theory Problems]]
 {{MAA Notice}}

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

Latest revision as of 08:47, 7 August 2014

Contents

Problem

Solutions

Solution 1

See also