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find value of above series
difficulty: moderate
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Prove that (m+n)! / m!n! Is an integer
ofc not using combiProve That product of n consecutive postive int. is divisible by n! (: VÊÇTØR :)
let a= x/y , b= p/q , c= l/m ( rational so ) x,y,p,q,l,m are integers
Given (x/y)+(p/q)2^(1/3) + (l/m)2^(2/3) = 0 , multiply both side by yqm
Let xqm= f , pym = g , lqy = h ofc f,g,h are integers
Use identity j^3 + k^3 + l^3 = 3jkl when j+k+l = 0
f^3 + 2g^3 + 4h^3 = 6fgh , now prove all f,g,h are divisble by 2
Put f=2f_1 , g= 2g_1 ,h = 2h_1 again come same equation in f_1, g_1 , h_1 so again it will divisible by 2 so f has infinite many 2 as a factor so f = 0 similarly g=h = 0 ( this is also called infinite decent method we have use to prove all are 0 )
Good question. Confidence booster when you know the source
One more question to prove eular theorem and Wilson theorem using ideas of special equal sets
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Again infinite decent with square of a odd int = 1( mod 8) will solved in a line
Use infinite decent and while using u should use square of a odd int = 1 (mod 8,4,2 )
Proof that this is equal to 0(mod p) which is looking obvious to prove
Try this RMO 2023, p_1 which is better than INMO 2024, p_3
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