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

Revision as of 08:10, 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

We proceed by induction.

Let ${a_{n}}$ be $7^{7^{n}}+1$ . The result holds for ${n=0}$ because ${a_0 = 2^3}$ is the product of $3$ primes.

Now we assume the result holds for ${n}$ . Note that ${a_{n}}$ satisfies the recursion ${a_{n+1}= (a_{n}-1)^{7}+1} = a_{n}\left(a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}\right).$

Since ${a_n - 1}$ is an odd power of ${7}$ , ${7(a_n-1)}$ is a perfect square. Therefore ${a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}}$ is a difference of squares and thus composite, i.e. it is divisible by ${2}$ primes. By assumption, ${a_n}$ is divisible by ${2n + 3}$ primes. Thus ${a_{n+1}}$ is divisible by ${2+ (2n + 3) = 2(n+1) + 3}$ primes as desired.

Solution 2

Notice that $7^{{7}^{k+1}}+1=(7+1) \frac{7^{7}+1}{7+1} \cdot \frac{7^{7^2} + 1}{7^{7^1} + 1} \cdots \frac{7^{7^{k+1}}+1}{7^{7^{k}}+1}$ . Therefore it suffices to show that $\frac{7^{7^{k+1}}+1}{7^{7^{k}}+1}$ is composite.

Let $x=7^{7^{k}}$ . The expression becomes $\frac{7^{7^{k+1}}+1}{7^{7^{k}}+1} = \frac{x^7 + 1}{x + 1},$ which is the shortened form of the geometric series $x^{6}-x^{5}+x^{4}-x^3 + x^2 - x+1$ . This can be factored as $(x^{3}+3x^{2}+3x+1)^{2}-7x(x^{2}+x+1)^{2}$ .

Since $x$ is an odd power of $7$ , $7x$ is a perfect square, and so we can factor this by difference of squares. Therefore, it is composite.

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

@@ Line 2: / Line 2: @@
 (''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.
-==Hint 1 of 3==
+==Solutions==
-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 ===
@@ Line 74: / Line 10: @@
 Now we assume the result holds for <math>{n}</math>. Note that <math>{a_{n}}</math> satisfies the [[recursion]]
+<cmath>{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).</cmath>
-<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.
@@ Line 83: / Line 18: @@
 Let <math>x=7^{7^{k}}</math>. The expression becomes
+<cmath>\frac{7^{7^{k+1}}+1}{7^{7^{k}}+1} = \frac{x^7 + 1}{x + 1},</cmath>
-<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.
+{{alternate solutions}}
 == See also ==

2007 USAMO (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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