求证:对于任意正整数n,总有无限个正整数K,使得n^+kn+1为完全平方数且若k>2 则k>√n
热心网友
1.对于任意正整数n,取k=2+s[(s+2)n+2],s=0,1,2,3,...,==n^2+kn+1=n^2+[2+s[s+2]n+2]]]n+1=[n+1+ns]^2.2.设n^2+kn+1=(n+a)^2,k2==a1==(k-2a)n=a^2-10==k2a==a^2-1≥n==a√n==k2√n.
热心网友
不会！
热心网友
求证:对于任意正整数n,总有无限个正整数K,使得n^+kn+1为完全平方数 且若k2 则k√n 令k=(m^2+2m)*n +2(m+1) ,m为自然数。则 n^2+kn+1=(m+1)^2*n^2 +2(m+1)n +1 = [(m+1)n +1]^2所以总有无数个正整数k ,使得n^2+kn+1为完全平方数因为k=(m^2+2m)*n +2(m+1)=[(m^2+2m)*n +2(m+1)-√n] +√n ={√[(m^2+2m)*√n]^2 - 1/[2√(m^2+2m)]}^2 +[2m+2 - 1/(4m^2+8m)] +√n ＞√n所以 k=(m^2+2m)*n +2m+2 ＞2 且k＞√n