1. Let n be a natural number and d be a positive divisor of 2n2. It is proved that N2+d is not a perfect square.
Topic 1953 Hungarian Mathematical Olympiad Topic 2.
It is proved that 2n2 = KD and k is a positive integer. If N2+d is the square of the integer x, then
k2x2=k2(n2+d)=n2(k2+2k)
But this is impossible, because both k2x2 and n2 are completely squared, and K2+2K < (k+ 1) 2 shows that K2+2K is not a square number.
Try to prove that the product of four consecutive natural numbers plus the arithmetic square root of 1 is still a natural number.
The title is 1962 and the final title is 1.
It is proved that the product of four continuous natural numbers can be expressed as
n(n+ 1)(n+2)(n+3)=(N2+3n)(N2+8n+2)
=(n2+3n+ 1)2- 1
Therefore, the product of four continuous natural numbers plus 1 is a complete square number, so we know that the conclusion of this problem is valid.
-
1. It is known that all terms are positive integers, and one of them is a complete square number. It is proved that this series must contain infinite complete squares.
Topic 1963 topic 2 of the tenth grade of the all-Russian mathematical Olympics. There are infinite arithmetic progression.
It is proved that the tolerance of this arithmetic progression is d, in which a = m2 (m ∈ n). therefore
a+(2km+dk2)d=(m+kd)2
For any k∈N, it is a term in arithmetic progression and a complete square number.
2. Find the largest complete square number, and after the last two digits are crossed out, you still get a complete square number (assuming that one of the crossed out two digits is non-zero).
Topic 1964 11th grade topic of all-Russian Mathematical Olympics 1.
Let n2 satisfy the condition that N2 = 100a2+b, where 0 < b < 100. So n > 10a, that is, n ≥ 10a+ 1. Therefore,
b=n2 100a2≥20a+ 1
This gives 20a+ 1 < 100, so a ≤ 4.
Empirical calculation shows that n = 4 1 satisfies the condition only when a = 4. If n > 4 1, N2-402 ≥ 422-402 > 100. Therefore, the maximum complete square number satisfying the condition of this question is 4 12.
-
1. Find all prime numbers P so that 4P2+ 1 and 6P2+ 1 are also prime numbers.
Topic 1964 ~ 1965 Polish Mathematical Olympiad Examination II 1.
P ≡ 1 (mod 5), 5 | 4p2+ 1 When p ≡ 2 (mod 5), 5 | 6p2+ 1. So there is only one solution to this problem, p = 5.
2. It is proved that there are infinite natural numbers A with the following properties: for any natural number n, z = n4+a is not a prime number.
The title of the 11th (1969) International Mathematical Olympiad is 1, which was provided by the former GDR.
It is proved that any integer m > 1 and natural number n exist.
n4+4m4=(n2+2m2)2-4m2n2
=(n2+2mn+2m2)(n2-2mn+2m2)
While N2+2mn+2m2 > N2-2mn+2m2.
=(n-m)2+m2≥m2> 1
So n4+4m4 is not a prime number. Let a = 4? 24,4? 34, ... you will get countless A's that meet the requirements.
-
1. If the natural number n makes 2n+ 1 and 3n+ 1 exactly square numbers, can 5n+3 be a prime number?
The topic is the 19th (1993) All-Russian Mathematical Olympiad, grade 9, test question 1.
Solution If 2n+ 1 = k2, 3n+ 1 = m2, then 5n+3 = 4 (2n+1)-(3n+1) = 4k2-m2 = (2k+m) (2k-m).
Because 5n+3 > (3n+1)+2 = m2+2 > 2m+1,2k-m ≠ 1 (otherwise 5n+3 = 2k+m = 2m+ 1). So 5n+3 =
2. What is the smallest natural number that can be expressed as the sum of 9 continuous natural numbers, 10 continuous natural numbers and 1 1 continuous natural numbers?
The 11th (1993) American Mathematics Invitational Tournament Topic 6.
Answer 495.
The sum of 9 consecutive integers is 9 times of the fifth number; The sum of 10 consecutive integers is 5 times the sum of items 5 and 6; The sum of 1 1 consecutive integers is 1 1 times of item 6, so the natural number that meets the requirements of the topic must be divisible by 9 and 5, 1 1, and this number is at least 495.
And 495 = 51+52+…+59 = 45+46+…+54 = 40+41+…+50.
3.02 1 Try to determine the largest positive integer A by the following properties: If all positive integers from 100 1 to 2000 are arranged in any order, we can find out the continuous 10 items from them, so that the sum of these 10 items is greater than or equal to a. 。
The 1st (1992) Mathematical Olympiad in Taipei, China, Question 6.
Solve any permutation, and the sum is 1001+1002+…+2000 =1500500, and divide it into100 segments, with each segment having10 and at least one segment.
A≥ 15005
On the other hand, the arrangement of 100 1 ~ 2000 is as follows:
2000 100 1 1900 1 10 1 1800
120 1 1700 130 1 1600 140 1
1999 1002 1899 1 102 1799
1202 1699 1302 1599 1402
… … … … … …
190 1 1 100 180 1 1200 170 1
1300 160 1 1400 150 1 1300
Please remember, the above arrangement is
a 1，a2，…，a2000
(The number of row I and column J in the table is 10 in this series (I- 1)+J, 1≤i≤20, 1≤j≤ 10).
Let Si = AI+AI+1+…+AI+9 (I =1,2, …, 190 1).
Then s 1 = 15005, S2 = 15004. It is easy to know that if I is an odd number, then Si =15005; If I is an even number, then Si = 15004.
To sum up, A = 15005.
-
What kind of natural number is 1? n？
32n+ 1-22n+ 1-6n
Is it a composite number?
The 24th (1990) All-Soviet Mathematical Olympiad, the fifth question in the eleventh grade.
The solution is 32n+1-22n+1-6n = (3n-2n) (3n+1+2n+1).
When n > L, 3n-2n > 1, 3n+1+2n+1>1,the original number is a composite number. When n = 1, the prime number is 13.
2. Prove that for any positive integer n, there are n consecutive positive integers, all of which are not integer powers of prime numbers.
The 30th (1989) International Mathematical Olympiad Question 5. This question was provided by Sweden.
Let a = (n+ 1)! Then a2+K (2 ≤ K ≤ N+ 1) can be divisible by K but not by k2 (because a2 can be divisible by k2 and K cannot be divisible by k2). If A2+K is an integer power pl of a prime number, then K = PJ (both L and J are positive integers), but A2 is divisible by p2j, so PJ+65438.
a2+k(2≤k≤n+ 1)
None of these n consecutive positive integers is an integer power of a prime number.
-
1. Find five different positive integers and make them pair prime numbers. The sum of any number n(n≤5) is a composite number.
10 Grade 2 1 Session (1987) The title of the All-Soviet Mathematical Olympics is 1.
This solution consists of n numbers.
ai=i？ n！ + 1，i= 1，2，…，n
The assembled unit meets the requirements.
Because the sum of any k number is
m？ n！ +k(m∈N，2≤k≤n)
Because n! = 1? 2? …? N is a multiple of k, so m? n！ +k is a multiple of k, so it is a composite number.
For any two numbers ai and AJ (I > J), if they have a common prime factor P, then P is also AI-AJ = (I-J) n! The prime factor of, because 0 < I-j < n, so p is also n! But ai and n! Mutual prime, so ai and aj can't have the same prime factor P, that is, ai and aj(i≠j) are mutually prime. Let n = 5, then we can get a set of numbers that meet the conditions: 12 1, 24 1, 36 1, 488.
Let the positive integer d not be equal to 2,5, 13. It is proved that two different elements A and B can be found in the set {2,5, 13, d}, so that AB- 1 is not a complete square number.
The title of the 27th (1986) International Mathematical Olympiad is 1. This title was provided by the former Federal Republic of Germany.
It is proved that at least one of 2D- 1, 5D- 1, 13D- 1 is not a complete square number. By reducing absurdity, let
5d- 1=x2 ( 1)
5d- 1=y2 (2)
13d- 1=z2 (3)
Where x, y and z are positive integers.
From the formula (1), X is odd, so we say X = 2n- 1. Substitute 2D- 1 = (2n- 1) 2, that is.
d=2n2-2n+ 1 (4)
Equation (4) shows that d is also an odd number.
Therefore, from (2) and (3), y and z are even numbers. Let Y = 2p and Z = 2q, substitute (2) and (3) and divide by 4.
2d=q2-p2=(q+p)(q-p)
Since 2d is even, that is, Q2-P2 is even, so both P and Q are even or odd, so both Q+P and Q-P are even, that is, 2d is a multiple of 4, so D is even. This contradicts that D is an odd number, so the proposition is correct.
-
1. If a natural number is a prime number and its numbers are exchanged at will, the number obtained is still a prime number, then such a number is called an absolute prime number. Prove that an absolute prime number cannot exceed three different numbers.
Eighth grade of the 18th (1984) All-Soviet Mathematical Olympiad.
If there are more than three different numbers, these numbers can only be 1, 3, 7 and 9. It is not difficult to verify 1379, 3 179, 9 137, 79 13, 1397, 3 138. The remainder is 0, 1, 2, 3, 4, 5 and 6 respectively. Therefore, for any natural number m, 104×M and the above seven four-digit numbers are added separately, and at least one of the sum obtained can be divisible by 7, so the number including 1, 3, 7, 9 is not an absolute prime number.
2. It is proved that if both p and P+2 are prime numbers greater than 3, then 6 is a factor of P+ 1.
The 5th (1973) Canadian Mathematical Olympiad, question 3.
It is proved that 2 is a factor of P+ 1 because p is an odd number.
Because P, P+ 1 and P+2 are divisible by 3, P+ 1 can be divisible by 3.
So 6 is a factor of P+ 1