We are mathematician persons must thank you Who invented the following problems 1 find n≥2, such that 48n^4+1 is a perfect square. 2 give 1,2,2,3,3,3,4,4,4,4,… What is the 5000th term of this sequence? 3 give m,n∈N,where m>n and g(x) be a function Satisfying [g(x)]^m=[g(x)]^n, herein[g(x)]^m Means the function g(x) applied m times as in g(g(g(g(…(g(x) ) )…) m times and not the Power of g(x), prove that g(x) is one-to-one If and only if g(x) is onto-function My conjecture There is infinitely n∈N such that n is the sum of two Perfect square two forms Solution 1 since 48n^4+1=3(2n)^4+1 Assume that there exists k ∈N such that 3(2n)^4+1=k^2 (2n)^4=((k-1)(k+1))/3 =(6m(6m+2))/3:(k-1),k,(k+1) =(3m+1)4m We see that (3m+1) is even when m is odd So, there is a contradiction. Therefore, there is no n>1 such that 48n^4+1 is a perfect square. 2 Hint: use formula 1+2+3+…+n=(n(n+1))/2 3 proof give y=g(x) and Give g(g(g(…(g(x)))…) m-times = g(g(…(g(x) )…) n-times (→) give g is one-to-one mean that If x_1, x_2∈D_g, g(x_1)= g(x_2), thenx_1=x_2 From g(g(g(…(g(x)))…) m-times = g(g(…(g(x) )…) n-times We have g(g(…(g(x) )…) = g(…(g(x) )…) And since m>n do like this, will obtain that g(g(g(…(g(x)))…) (m-n)-times =x It result in g is the inverse of it That is g(g(g(…(g(x)))…) = g(g(…(g(x) )…) = g(…(g(x) )…) ⋮ =g(x)=x We see that g is onto-function (←) give g is onto-function mean that For each y_1∈R_g, there exists unique x_1∈D_g such that g(x_1 )=y_1 Please show by yourself. If this proof is false, please help me.
Posted on: Thu, 13 Mar 2014 09:01:25 +0000
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