2007 USAMO Problems/Problem 5: Difference between revisions

Latest revision as of 14:30, 9 October 2025

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.

Solution 2

We prove by induction on $n$ that $7^{7^n} + 1$ is the product of at least $2n + 3$ (not necessarily distinct) primes.

For $n = 0$ , we have $7^{7^0} + 1 = 7 + 1 = 8 = 2^3$ , which has $3 = 2(0) + 3$ prime factors. Thus, the base case holds.

Assume that for some $n \ge 0$ , the number $7^{7^n} + 1$ is the product of at least $2n + 3$ primes. We must show that $7^{7^{n+1}} + 1$ has at least $2n + 5$ prime factors.

Let $a = 7^n$ . Since $a$ is odd, we can apply the Aurifeuillian factorization: $7^{7a} + 1 = (7^a + 1)\big(7^{3a} + 3\cdot7^{2a} + 3\cdot7^a + 1\big)\big(7^{\tfrac{5a}{2} + \tfrac{1}{2}} + 7^{\tfrac{3a}{2} + \tfrac{1}{2}} + 7^{\tfrac{a}{2} + \tfrac{1}{2}}\big)$ .

Each factor is an integer greater than 1, so the factorization is nontrivial.

The first factor is $7^a + 1 = 7^{7^n} + 1$ , which by the inductive hypothesis has at least $2n + 3$ prime factors. It remains to show that the quotient $\frac{7^{7a} + 1}{7^a + 1}$ is composite.

From the factorization above, we have $\frac{7^{7a} + 1}{7^a + 1} = \big(7^{3a} + 3\cdot7^{2a} + 3\cdot7^a + 1\big)\big(7^{\tfrac{5a}{2} + \tfrac{1}{2}} + 7^{\tfrac{3a}{2} + \tfrac{1}{2}} + 7^{\tfrac{a}{2} + \tfrac{1}{2}}\big)$ ,

which is clearly the product of two integers greater than 1, and thus composite.

Therefore, $\frac{7^{7a} + 1}{7^a + 1}$ is always composite, and hence $7^{7a} + 1$ has at least two more prime factors than $7^a + 1$ .

By the inductive hypothesis, $7^a + 1$ has at least $2n + 3$ prime factors, so $7^{7a} + 1$ has at least $(2n + 3) + 2 = 2n + 5$ prime factors.

By induction, $7^{7^n} + 1$ is the product of at least $2n + 3$ (not necessarily distinct) primes for all nonnegative integers $n$ .

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

@@ Line 1: / Line 1: @@
 == 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) primes.
+==Solutions==
-== Solution ==
+=== Solution 1 ===
+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.
-{{USAMO newbox|year=2007|num-b=4|num-a=6}}
+=== Solution 2===
+We prove by induction on <math>n</math> that <math>7^{7^n} + 1</math> is the product of at least <math>2n + 3</math> (not necessarily distinct) primes.
+For <math>n = 0</math>, we have <math>7^{7^0} + 1 = 7 + 1 = 8 = 2^3</math>, which has <math>3 = 2(0) + 3</math> prime factors.
+Thus, the base case holds.
+Assume that for some <math>n \ge 0</math>, the number <math>7^{7^n} + 1</math> is the product of at least <math>2n + 3</math> primes.
+We must show that <math>7^{7^{n+1}} + 1</math> has at least <math>2n + 5</math> prime factors.
+Let <math>a = 7^n</math>. Since <math>a</math> is odd, we can apply the Aurifeuillian factorization:
+<math>7^{7a} + 1 = (7^a + 1)\big(7^{3a} + 3\cdot7^{2a} + 3\cdot7^a + 1\big)\big(7^{\tfrac{5a}{2} + \tfrac{1}{2}} + 7^{\tfrac{3a}{2} + \tfrac{1}{2}} + 7^{\tfrac{a}{2} + \tfrac{1}{2}}\big)</math>.
+Each factor is an integer greater than 1, so the factorization is nontrivial.
+The first factor is <math>7^a + 1 = 7^{7^n} + 1</math>, which by the inductive hypothesis has at least <math>2n + 3</math> prime factors.
+It remains to show that the quotient
+<math>\frac{7^{7a} + 1}{7^a + 1}</math>
+is composite.
+From the factorization above, we have
+<math>\frac{7^{7a} + 1}{7^a + 1} = \big(7^{3a} + 3\cdot7^{2a} + 3\cdot7^a + 1\big)\big(7^{\tfrac{5a}{2} + \tfrac{1}{2}} + 7^{\tfrac{3a}{2} + \tfrac{1}{2}} + 7^{\tfrac{a}{2} + \tfrac{1}{2}}\big)</math>,
+which is clearly the product of two integers greater than 1, and thus composite.
+Therefore, <math>\frac{7^{7a} + 1}{7^a + 1}</math> is always composite, and hence <math>7^{7a} + 1</math> has at least two more prime factors than <math>7^a + 1</math>.
+By the inductive hypothesis, <math>7^a + 1</math> has at least <math>2n + 3</math> prime factors, so <math>7^{7a} + 1</math> has at least <math>(2n + 3) + 2 = 2n + 5</math> prime factors.
-Let <math>\displaystyle{a_{n}}</math> be <math>7^{7^{n}}+1</math>. We prove the result by induction.
+By induction, <math>7^{7^n} + 1</math> is the product of at least <math>2n + 3</math> (not necessarily distinct) primes for all nonnegative integers <math>n</math>.
-The result holds for <math>\displaystyle{n=0}</math> because <math>\displaystyle{a_0 = 2^3}</math> is the product of <math>\displaystyle{3}</math> primes. Now we assume the result holds for <math>\displaystyle{n}</math>. Note that <math>\displaystyle{a_{n}}</math> satisfies the recursion
+{{alternate solutions}}
-<math>\displaystyle{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>.
+== See also ==
+* <url>viewtopic.php?t=145849 Discussion on AoPS/MathLinks</url>
+{{USAMO newbox|year=2007|num-b=4|num-a=6}}
-Since <math>\displaystyle{a_n - 1}</math> is an odd power of <math>\displaystyle{7}</math>, <math>\displaystyle{7(a_n-1)}</math> is a perfect square. Therefore <math>\displaystyle{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>\displaystyle{2}</math> primes. By assumption, <math>\displaystyle{a_n}</math> is divisible by <math>\displaystyle{2n + 3}</math> primes. Thus <math>\displaystyle{a_{n+1}}</math> is divisible by <math>\displaystyle{2+ (2n + 3) = 2(n+1) + 3}</math> primes as desired.
+[[Category:Olympiad Number Theory Problems]]
+{{MAA Notice}}

2007 USAMO (Problems • Resources)
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