Difference between revisions of "2008 USAMO Problems/Problem 1"
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== Problem == | == Problem == | ||
− | (''Titu Andreescu'') | + | (''Titu Andreescu'') Prove that for each positive integer <math>n</math>, there are pairwise relatively prime integers <math>k_0, k_1 \dotsc, k_n</math>, all strictly greater than 1, such that <math>k_0 k_1 \dotsm k_n -1</math> is the product of two consecutive integers. |
− | Prove that for each positive integer <math>n</math>, there are pairwise relatively prime integers <math>k_0, k_1 \dotsc, k_n</math>, all strictly greater than 1, such that <math>k_0 k_1 \dotsm k_n -1</math> is the product of two consecutive integers. | ||
− | |||
== Solutions == | == Solutions == | ||
+ | |||
=== Solution 1 === | === Solution 1 === | ||
We will prove the problem for each nonnegative integer <math>n</math>. We wish to show that | We will prove the problem for each nonnegative integer <math>n</math>. We wish to show that | ||
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&= (a+1)^4 + (a+1)^2 + 1 . | &= (a+1)^4 + (a+1)^2 + 1 . | ||
\end{align*} </cmath> | \end{align*} </cmath> | ||
− | This completes the induction. | + | This completes the induction. |
=== Solution 2 === | === Solution 2 === | ||
'''Lemma.''' If <math>p</math> is prime such that <math>p\equiv 1 \pmod{3}</math>, there exists a residue <math>r</math> such that <math>p \mid r^2+r+1</math>. | '''Lemma.''' If <math>p</math> is prime such that <math>p\equiv 1 \pmod{3}</math>, there exists a residue <math>r</math> such that <math>p \mid r^2+r+1</math>. | ||
− | ''Proof.'' Let <math>g</math> be a multiplicative generator of the nonzero integers mod | + | ''Proof.'' Let <math>g</math> be a multiplicative generator of the nonzero integers mod p. Set <math>r \equiv g^{(p-1)/3}</math>. Then <math>r-1 \not\equiv 0 \pmod{p}</math>, but <math>r^3-1 \equiv 0 \pmod{p}</math>, so <math>\frac{r^3-1}{r-1} \equiv r^2 + r+1 \equiv 0 \pmod{p}</math>. |
+ | |||
+ | '''End Lemma''' | ||
By [[Dirichlet's Theorem]], there are infinitely many primes congruent to 1 (mod 3). Let <math>p_0, \dotsc, p_n</math> be <math>(n+1)</math> such primes, and let <math>r_0, \dotsc, r_n</math> be respective residues as described in the lemma. By the [[Chinese Remainder Theorem]], there is a positive integer <math>a</math> that satisfies the relation | By [[Dirichlet's Theorem]], there are infinitely many primes congruent to 1 (mod 3). Let <math>p_0, \dotsc, p_n</math> be <math>(n+1)</math> such primes, and let <math>r_0, \dotsc, r_n</math> be respective residues as described in the lemma. By the [[Chinese Remainder Theorem]], there is a positive integer <math>a</math> that satisfies the relation | ||
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for each integer <math>i \in [0,n]</math>. Then | for each integer <math>i \in [0,n]</math>. Then | ||
<cmath> p_0 \dotsc p_n \mid a^2+a+1 . </cmath> | <cmath> p_0 \dotsc p_n \mid a^2+a+1 . </cmath> | ||
− | Now, for <math>1 \le i \le n</math>, take <math>k_i</math> to be the greatest power of <math>p_i</math> that divides <math>a^2 + a +1</math>, and let <math>k_0 = (a^2+a+1)/(k_1 \dotsm k_n)</math>. Since all the <math>k_i</math> are pairwise relatively prime and are greater than 1, we are done. | + | Now, for <math>1 \le i \le n</math>, take <math>k_i</math> to be the greatest power of <math>p_i</math> that divides <math>a^2 + a +1</math>, and let <math>k_0 = (a^2+a+1)/(k_1 \dotsm k_n)</math>. Since all the <math>k_i</math> are pairwise relatively prime and are greater than 1, we are done. |
=== Solution 3 === | === Solution 3 === | ||
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Define <math>P</math> to be the greatest possible cardinality of a relatively prime partition for that number. | Define <math>P</math> to be the greatest possible cardinality of a relatively prime partition for that number. | ||
− | Lemma 1 | + | '''Lemma 1.''' All cardinalities of the relatively prime partitions of a number up to <math>P</math> can be attained. |
− | Proof | + | ''Proof.'' <math>P = P - 0</math> satisfies the properties. For any <math>P - n</math> which satisfies the properties, we can take any of the 2 <math>P - n</math> numbers and multiply them together. Because they are both relatively prime to all the other numbers, their product is relatively prime to all the other numbers as well, and that results in <math>P - n - 1</math> numbers which satisfy the conditions, unless <math>P - n = 1</math>, because there is only one number left. Therefore, all numbers of numbers of relatively prime factors from <math>1</math> to <math>P - 1</math> are attainable, if <math>P</math> is attainable as well. |
− | <math>P = P - 0</math> satisfies the properties. For any <math>P - n</math> which satisfies the properties, we can take any of the 2 <math>P - n</math> numbers and multiply them together. Because they are both relatively prime to all the other numbers, their product is relatively prime to all the other numbers as well, and that results in <math>P - n - 1</math> numbers which satisfy the conditions, unless <math>P - n = 1</math>, because there is only one number left. Therefore, all numbers of numbers of relatively prime factors from <math>1</math> to <math>P - 1</math> are attainable, if <math>P</math> is attainable as well. | ||
− | End Lemma | + | '''End Lemma''' |
− | Lemma 2 | + | '''Lemma 2.''' <math>a^2 + a + 1</math> can have arbitrarily many prime factors for some value of <math>a</math>. |
− | Proof | + | ''Proof.'' Let <math>f(x) = x^2 + x + 1</math>. Let <math>a^2 + a + 1 = k</math>. Then <math>f(a + k) = a^2 + 2ak + k^2 + a + k + 1 = 2ak + k^2 + 2k</math>. |
− | Let <math>f(x) = x^2 + x + 1</math>. Let <math>a^2 + a + 1 = k</math>. Then <math>f(a + k) = a^2 + 2ak + k^2 + a + k + 1 = 2ak + k^2 + 2k</math>. | ||
After factoring, we get <math>k(k + 2 + 2a) = (a^2 + a + 1)(a^2 + 3a + 3)</math>. | After factoring, we get <math>k(k + 2 + 2a) = (a^2 + a + 1)(a^2 + 3a + 3)</math>. | ||
<math>a^2 + 3a + 3 \geq a^2 + a + 1</math>, because <math>a</math> is strictly positive. | <math>a^2 + 3a + 3 \geq a^2 + a + 1</math>, because <math>a</math> is strictly positive. | ||
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<math>(a^2 + a + 1)(a^2 + 3a + 3)</math> therefore must contain prime factors not in <math>a^2 + a + 1</math>. Upon repeating this an arbitrary number of times, we get a number of the form <math>a^2 + a + 1</math> which has arbitrarily many distinct prime factors. | <math>(a^2 + a + 1)(a^2 + 3a + 3)</math> therefore must contain prime factors not in <math>a^2 + a + 1</math>. Upon repeating this an arbitrary number of times, we get a number of the form <math>a^2 + a + 1</math> which has arbitrarily many distinct prime factors. | ||
− | End Lemma | + | '''End Lemma''' |
Then <math>a(a + 1) + 1 = a^2 + a + 1</math> would not have <math>n + 1 = P</math> elements in the relatively prime partition for some value of n, and any value of a. | Then <math>a(a + 1) + 1 = a^2 + a + 1</math> would not have <math>n + 1 = P</math> elements in the relatively prime partition for some value of n, and any value of a. | ||
It is possible to choose a value of <math>a</math> such that <math>a^2 + a + 1</math> has arbitrarily many unique prime factors, by our second Lemma, and so it is possible for <math>P</math> to be arbitrarily high. By our first Lemma, all numbers up to P are possible values for the cardinality of some relatively prime partition, and so there always exists some number that has an arbitrary number of elements in a relatively prime partition. Because <math>a^2 + a + 1 = a(a + 1) + 1</math>, <math>a(a + 1)</math> is the product of 2 consecutive integers, we see that the given statement is true because if <math>k_0k_1...k_n = a(a + 1) + 1</math>, then <math>k_0k_1...k_n - 1 = a(a + 1)</math>. | It is possible to choose a value of <math>a</math> such that <math>a^2 + a + 1</math> has arbitrarily many unique prime factors, by our second Lemma, and so it is possible for <math>P</math> to be arbitrarily high. By our first Lemma, all numbers up to P are possible values for the cardinality of some relatively prime partition, and so there always exists some number that has an arbitrary number of elements in a relatively prime partition. Because <math>a^2 + a + 1 = a(a + 1) + 1</math>, <math>a(a + 1)</math> is the product of 2 consecutive integers, we see that the given statement is true because if <math>k_0k_1...k_n = a(a + 1) + 1</math>, then <math>k_0k_1...k_n - 1 = a(a + 1)</math>. | ||
+ | |||
+ | === Solution 4 === | ||
+ | Write the relation to be proved as | ||
+ | <cmath>4k_0k_1\cdots k_n = 4a(a + 1) + 4 = (2a + 1)^2 + 3.</cmath> | ||
+ | There are infinitely many primes for which <math>-3</math> is a quadratic residue. Let <math>2 < p_0 < p_1 < \cdots < p_n</math> be such primes. Using the Chinese Remainder Theorem to specify <math>a</math> modulo <math>p_n</math>, we can find an integer <math>a</math> such that <math>(2a + 1)^2 + 3 = 4p_0p_1\cdots p_nm</math> for some positive integer <math>m</math>. Grouping the factors of <math>m</math> appropriately with the <math>p_i</math>'s, we obtain <math>(2a + 1)^2 + 3 = 4k_0k_1\cdots k_n</math> with <math>k_i</math> pairwise relatively prime. We then have <math>k_0k_1\cdots k_n - 1 = a(a + 1)</math>, as desired. | ||
+ | |||
+ | === Solution 5 === | ||
+ | We are supposed to show that for every positive integer <math>n</math>, there is a positive integer <math>x</math> such that <math>x(x + 1) + 1 = x^2 + x + 1</math> has at least <math>n</math> distinct prime divisors. We can actually prove a more general statement. | ||
+ | |||
+ | '''Claim.''' Let <math>P(x) = a_dx^d + \cdots + a_1x + 1</math> be a polynomial of degree <math>d\geq 1</math> with integer coefficients. Then for every positive integer <math>n</math>, there is a positive integer <math>x</math> such that <math>P(x)</math> has at least <math>n</math> distinct prime divisors. | ||
+ | |||
+ | The proof follows from the following two lemmas. | ||
+ | |||
+ | '''Lemma 1.''' The set | ||
+ | <cmath>Q = \{p\mid p\text{ a prime for which there is an integer }x\text{ such that }p\text{ divides }P(x)\}</cmath> | ||
+ | is infinite. | ||
+ | |||
+ | ''Proof.'' The proof is analogous to Euclid's proof that there are infinitely many primes. Namely, if we assume that there are only finitely many primes <math>p_1, p_2, \ldots, p_k</math> in <math>Q</math>, then for each integer <math>m</math>, <math>P(mp_1p_2\cdots p_k)</math> is an integer with no prime factors, which must equal <math>1</math> or <math>-1</math>. However, since <math>P</math> has degree <math>d\geq 1</math>, it takes each of the values <math>1</math> and <math>-1</math> at most <math>d</math> times, a contradiction. | ||
+ | |||
+ | '''End Lemma''' | ||
+ | |||
+ | '''Lemma 2.''' Let <math>p_1, p_2, \ldots, p_n</math>, <math>n\geq 1</math> be primes in <math>Q</math>. Then there is a positive integer <math>x</math> such that <math>P(x)</math> is divisible by <math>p_1p_2\cdots p_n</math>. | ||
+ | |||
+ | ''Proof.'' For <math>i = 1, 2, \ldots, n</math>, since <math>p_i\in Q</math> we can find an integer <math>c_i</math> such that <math>P(x)</math> is divisible by <math>p_i</math> whenever <math>x\equiv c_i\pmod{p_i}</math>. By the Chinese Remainder Theorem, the system of <math>n</math> congruences <math>x\equiv c_i\pmod{p_i}</math>, <math>i = 1, 2, \ldots, n</math> has positive integer solutions. For every positive integer <math>x</math> that solves this system, <math>P(x)</math> is divisible by <math>p_1p_2\cdots p_n</math>. | ||
+ | |||
+ | '''End Lemma''' | ||
+ | |||
+ | == Solution 6 == | ||
+ | Very similar to some above solutions. | ||
+ | |||
+ | Let <math>k_{0}k_{1}\dots k_{n}-1=a(a+1)</math>. We wish to show this equation has a solution <math>(a,k_{0},\dots,k_{n})\in\mathbb{Z}^{n+2}</math> for any <math>n\in\mathbb{N}</math>. If <math>a^{2}+a+1</math> has at least <math>n+1</math> distinct prime factors, we can set <math>a^{2}+a+1=p_{1}^{q_{1}}p_{2}^{q_{2}}\dots p_{m}^{q_{m}}</math> for some <math>m\geq n</math> where <math>p_{1},\dots p_{m}</math> are distinct primes. Then, we let <math>k_{i}=p_{i+1}^{q_{i+1}}</math> for all <math>0\leq i\leq n-1</math> and set <math>k_{n}=p_{n+1}^{q_{n+1}}\dots p_{m}^{q_{m}}</math>. Thus, it remains to show <math>a^{2}+a+1</math> can have at least <math>n+1</math> distinct prime factors. We will do this by showing it can have arbitrarily many. | ||
+ | |||
+ | Using an induction-type argument (it isn't exactly induction though), we first show that <math>a^{2}+a+1</math> can have <math>1</math> prime factor by setting <math>a=2</math>. Now, assume <math>a^{2}+a+1</math> has <math>n</math> distinct prime factors. We wish to show there exists some <math>b</math> such that <math>b^{2}+b+1</math> has more than <math>n</math> distinct prime factors (we need not show it has <math>n+1</math>, since we only wish to demonstrate that the number of prime factors can be arbitrarily large rather than any integer). Set <math>b=a^{2}</math>. Then | ||
+ | <cmath>\begin{align*} | ||
+ | b^{2}+b+1&=(b+1)^{2}-b\\ | ||
+ | &=(a^{2}+1)^{2}-a^{2}\\ | ||
+ | &=(a^{2}+a+1)(a^{2}-a+1). | ||
+ | \end{align*}</cmath> | ||
+ | Suppose <math>p</math> is a positive integer dividing both <math>a^{2}+a+1</math> and <math>a^{2}-a+1</math>. Then <math>p\mid(a^{2}+a+1)-(a^{2}-a+1)=2a</math>, and because <math>(a^{2}+a+1)</math> is always odd, we have <math>2\nmid p</math> so <math>p\mid a</math>. But this yields <math>p\mid\gcd(a,a^{2}+a+1)=1</math>, so <math>p=1</math> and thus <math>a^{2}+a+1</math> and <math>a^{2}-a+1</math> are coprime. Therefore, as one can make <math>a^{2}-a+1>1</math>, we see that <math>a^{2}-a+1</math> has at least one prime factor which does not divide <math>a^{2}+a+1</math> (which has <math>n</math> distinct prime factors). Therefore, <math>b^{2}+b+1</math> has more than <math>n</math> distinct prime factors. | ||
{{alternate solutions}} | {{alternate solutions}} | ||
− | == | + | == See also == |
− | + | * <url>viewtopic.php?t=202909 Discussion on AoPS/MathLinks</url> | |
− | + | {{USAMO newbox|year=2008|before=First Question|num-a=2}} | |
[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 03:38, 17 August 2019
Contents
Problem
(Titu Andreescu) Prove that for each positive integer , there are pairwise relatively prime integers , all strictly greater than 1, such that is the product of two consecutive integers.
Solutions
Solution 1
We will prove the problem for each nonnegative integer . We wish to show that for some integer . We induct on . For our base case, , we may let be positive integer.
For the inductive step, suppose that are pairwise relatively prime integers such that for some integer . Let . Evidently, . Also, Since is odd and relatively prime to , it follows that and are relatively prime, so is relatively prime to each of . Finally, This completes the induction.
Solution 2
Lemma. If is prime such that , there exists a residue such that .
Proof. Let be a multiplicative generator of the nonzero integers mod p. Set . Then , but , so .
End Lemma
By Dirichlet's Theorem, there are infinitely many primes congruent to 1 (mod 3). Let be such primes, and let be respective residues as described in the lemma. By the Chinese Remainder Theorem, there is a positive integer that satisfies the relation for each integer . Then Now, for , take to be the greatest power of that divides , and let . Since all the are pairwise relatively prime and are greater than 1, we are done.
Solution 3
Firstly, we see that there are numbers . Since , there are at least 2 values of . Define a relatively prime partition to be a set of relatively prime numbers such that their product is equal to some natural number.
Define to be the greatest possible cardinality of a relatively prime partition for that number.
Lemma 1. All cardinalities of the relatively prime partitions of a number up to can be attained.
Proof. satisfies the properties. For any which satisfies the properties, we can take any of the 2 numbers and multiply them together. Because they are both relatively prime to all the other numbers, their product is relatively prime to all the other numbers as well, and that results in numbers which satisfy the conditions, unless , because there is only one number left. Therefore, all numbers of numbers of relatively prime factors from to are attainable, if is attainable as well.
End Lemma
Lemma 2. can have arbitrarily many prime factors for some value of .
Proof. Let . Let . Then . After factoring, we get . , because is strictly positive. Let their GCD=g. , and so , but which is strictly odd, so g=1. therefore must contain prime factors not in . Upon repeating this an arbitrary number of times, we get a number of the form which has arbitrarily many distinct prime factors.
End Lemma
Then would not have elements in the relatively prime partition for some value of n, and any value of a. It is possible to choose a value of such that has arbitrarily many unique prime factors, by our second Lemma, and so it is possible for to be arbitrarily high. By our first Lemma, all numbers up to P are possible values for the cardinality of some relatively prime partition, and so there always exists some number that has an arbitrary number of elements in a relatively prime partition. Because , is the product of 2 consecutive integers, we see that the given statement is true because if , then .
Solution 4
Write the relation to be proved as There are infinitely many primes for which is a quadratic residue. Let be such primes. Using the Chinese Remainder Theorem to specify modulo , we can find an integer such that for some positive integer . Grouping the factors of appropriately with the 's, we obtain with pairwise relatively prime. We then have , as desired.
Solution 5
We are supposed to show that for every positive integer , there is a positive integer such that has at least distinct prime divisors. We can actually prove a more general statement.
Claim. Let be a polynomial of degree with integer coefficients. Then for every positive integer , there is a positive integer such that has at least distinct prime divisors.
The proof follows from the following two lemmas.
Lemma 1. The set is infinite.
Proof. The proof is analogous to Euclid's proof that there are infinitely many primes. Namely, if we assume that there are only finitely many primes in , then for each integer , is an integer with no prime factors, which must equal or . However, since has degree , it takes each of the values and at most times, a contradiction.
End Lemma
Lemma 2. Let , be primes in . Then there is a positive integer such that is divisible by .
Proof. For , since we can find an integer such that is divisible by whenever . By the Chinese Remainder Theorem, the system of congruences , has positive integer solutions. For every positive integer that solves this system, is divisible by .
End Lemma
Solution 6
Very similar to some above solutions.
Let . We wish to show this equation has a solution for any . If has at least distinct prime factors, we can set for some where are distinct primes. Then, we let for all and set . Thus, it remains to show can have at least distinct prime factors. We will do this by showing it can have arbitrarily many.
Using an induction-type argument (it isn't exactly induction though), we first show that can have prime factor by setting . Now, assume has distinct prime factors. We wish to show there exists some such that has more than distinct prime factors (we need not show it has , since we only wish to demonstrate that the number of prime factors can be arbitrarily large rather than any integer). Set . Then Suppose is a positive integer dividing both and . Then , and because is always odd, we have so . But this yields , so and thus and are coprime. Therefore, as one can make , we see that has at least one prime factor which does not divide (which has distinct prime factors). Therefore, has more than distinct prime factors.
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=202909 Discussion on AoPS/MathLinks</url>
2008 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.