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标题: Math Problem: Integer [打印本页]

作者: t1e1man 时间: 2006-12-1 01:46
标题: Math Problem: Integer

28th Annual University of Maryland-High School Mathematics Competition

Part II; November 29, 2006

Question 3: Prove that there are no integers m,n>=1 such that

sqrt(m+sqrt(m+sqrt(m+...+sqrt(m))))=n

where there are 2006 square root signs.

Question 5: Each positive integer is assigned one of three colors. Show that there exist disticnt positive integers x, y such that x and y have the same color and abs(x-y) is a perfect square.

本贴由[t1e1man]最后编辑于：2006-11-30 17:48:16

作者: xlxk 时间: 2006-12-1 16:39
标题: 回复：Math Problem: Integer

第一题：
sqrt(m+sqrt(m+...+sqrt(m+sqrt(m))))=n 如果m，n都是整数=>
sqrt(m+sqrt(m+...+sqrt(m+sqrt(m)))) （2005个开方号）是一个整数
sqrt(m+sqrt(m+...+sqrt(m+sqrt(m)))) （2004个开方号）是一个整数
......
sqrt(m+sqrt(m)) 是一个整数
sqrt(m) 是一个整数
假设m＝M^2
M
第二题
3^2 + 4^2 = 5^2
5^2 + 12^2 = 13^2
6^2 + 8^2 = 10^2
7^2 + 24^2 = 25^2
8^2 + 15^2 = 17^2
9^2 + 12^2 = 15^2
10^2 + 24^2 = 26^2
12^2 + 16^2 = 20^2
15^2 + 20^2 = 25^2

suppose the three colors are color A, B, and C and 1 is in color A:
then 17^2+1 can't be in A, suppose 17^2+1 is in B, for the first step we have:

Step 1-- 8^2+1 and 15^2+1 must be in C
A: 1
B: 17^2+1
C: 8^2+1 15^2+1

Step 2-- 25^2+1 can't be in C since 15^2+1 is in C. So 20^2+1 must be in C:
A: 1
B: 17^2+1 25^2+1
C: 8^2+1 15^2+1 20^2+1

Step 3-- 10^2+1 can't be in C since 8^2+1 is in C:
A: 1
B: 17^2+1 25^2+1 10^2+1
C: 8^2+1 15^2+1 20^2+1

Step 4-- 26^2+1 must be in C since 10^2+1 is in B. So 24^2+1 must be in B
A: 1
B: 17^2+1 25^2+1 10^2+1 24^2+1
C: 8^2+1 15^2+1 20^2+1 26^2+1

but (25^2+1)-(24^2-1)=7^2

本贴由[xlxk]最后编辑于：2006-12-1 13:23:49

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