Source: Romanian IMO TST 2006, day 5, problem 1
Tags: number theory unsolved, number theory
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Let n be a positive integer of the form 4 k + 1 , k ∈ N and A = { a 2 + n b 2 ∣ a , b ∈ Z } . Prove that there exist integers x , y such that x n + y n ∈ A and x + y ∉ A .
n=1 positive integer of the form 4 k + 1 . Prove that x + y ∈ A and x + y ∉ A .
I think that in Romania 0 is not a positive integer so k ≠ 0 :
Already posted by me last year...
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Moldavian&t=46004
A little problem with solution given there.
maxal wrote:
because u^2 + n*v^2 \equiv 0\pmod p has no solutions
case u\equiv v\equiv0 failes entire solution.
A very easy solution: x=2^{n+1}+2 , y=2^n+1 , x+y\equiv3\mod 4 , hence x+y\notin A . x^n+y^n=((2^n+1)^{2k+1})^2\in A .