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#1
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A Olympiad problem (original creation)
I have created many Olympiad problems, most of them for regional, state and National (Mexico) contests. This is not the one I'm proud the most, just one I find interesting and with a short solution.
(n,r) means nCr Find all positive integers n,z,k that satisfy: (n,0)^2+(n,1)^2+(n,2)^2+.....+(n,n)^2 = z^k |
#2
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Re: A Olympiad problem (original creation)
LHS = nC0*ncn + nC1*nC(n-1) + nC2*nC(n-2) +...+nCn*nC0
Suppose you have n girls and n boys to choose from to make a sports team but you must select n players . The total number of teams using n players is the lhs . It is also equivalent to 2nCn since we must select n players from 2n players . So z^k = (2nCn)^1 We have an immediate solution n=n , z=2nCn and k=1 . We need to show that the kth root of 2nCn is not an integer for k>=2 . Now we use the fact that there is always a prime number between n and 2n . This means that the prime factorization of 2nCn contains some prime number between n and 2n which is used only once . Therefore it cannot possibly be used k times (for k>=2) which is the lhs of the equation . The solutions are (n,z,k) = (n,2nCn ,1) |
#3
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Re: A Olympiad problem (original creation)
[ QUOTE ]
LHS = nC0*ncn + nC1*nC(n-1) + nC2*nC(n-2) +...+nCn*nC0 Suppose you have n girls and n boys to choose from to make a sports team but you must select n players . The total number of teams using n players is the lhs . It is also equivalent to 2nCn since we must select n players from 2n players . So z^k = (2nCn)^1 We have an immediate solution n=n , z=2nCn and k=1 . We need to show that the kth root of 2nCn is not an integer for k>=2 . Now we use the fact that there is always a prime number between n and 2n . This means that the prime factorization of 2nCn contains some prime number between n and 2n which is used only once . Therefore it cannot possibly be used k times (for k>=2) which is the lhs of the equation . The solutions are (n,z,k) = (n,2nCn ,1) [/ QUOTE ] Is there a solution that doesn't use Erdos's poem? |
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