Math Problem: Integer

t1e1man

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

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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.

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  本贴由[t1e1man]最后编辑于：2006-11-30 17:48:16  

xlxk

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 = 25^2www.ddhw.com 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+1www.ddhw.com 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 www.ddhw.com     本贴由[xlxk]最后编辑于：2006-12-1 13:23:49