RMM P1
Amy and Bob play the game. At the beginning, Amy writes down a positive integer on the board. Then the players take moves in turn, Bob moves first. On any move of his, Bob replaces the number on the blackboard with a number of the form , where a is a positive integer. On any move of hers, Amy replaces the number on the blackboard with a number of the form , where is a positive integer. Bob wins if the number on the board becomes . Amy wins if she can prevent the number from ever becoming .
Who will win?
Assumption: both Amy and Bob play optimally.
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1 solution
Let n be the initial number written on the board.From Lagrange's Theorem n = a 2 + b 2 + c 2 + d 2 . B choses a 2 .Now if A is writing a perfect square ( raising it to even power)so we are done or A writes a number of the form ( b 2 + c 2 + d 2 ) T 1 2 .Now B choses b 2 T 1 2 .Again A is writing a perfect square or a number of the form ( c 2 + d 2 ) T 2 2 .Now B choses c 2 T 2 2 and the number left is a perfect square so B wins after A moves.